Here are some questions on arithmetic progression and their solutions

**Question 1**: Which term of the A.P. 3,8,13 …is 78?

**Solution**: Here a_{n }= a + (n – 1) d = 78

a= 3, d = 8- 3 = 5

Therefore,

3 + (n -1) (5) = 78

(n-1) * 5 = 78 – 3 = 75

n – 1 = 75/5 = 15

n = 15 + 1 = 16

Hence a _{16 }(sixteenth term) is 78.

**Question 2**: Is – 150 a term of the series 11, 8, 5, 2,…?

**Solution**: Here, a = 11, d = 8-11 = -3. Let a_{n }= -150

Therefore,

a + (n-1) d = -150

11+ (n-1) (-3) = -150

-3 (n-1) = -150 – 11 = -161

(n-1) = + 161/3 = 53 2/3 which is not an integral number.

Since number of terms can never be a fraction

Hence, -150 is not a term of the given series.

**Question 3**: Find the 31^{st} term of an A.P. whose 11^{th} term is 38 and the 16^{th} term is 73.

**Solution:** Let a be the 1^{st} term and d the common difference.

Here a_{11 }= a + 10d = 38**….. (1)**

a_{16 }= a + 15d = 73…..** (2)**

Subtracting (2) from (1), we get

a + 10d – 1 – 15 d = 38 – 73

-5 = – 35

d = 7

Putting d= 7 in (1), we get

a + 10 * 7 = 38

a= 38 – 70 = – 32

a_{31}= a + 30 d

= -32 + 30 * 7

= -32 + 210 = 178

Hence, 31^{st} term is 178.

**Question 4**: Which term of the A.P. 3, 15, 27, 39 … will be 132 more than its 54^{th} term?

**Solution:** Given series is 3, 15, 27, 39…

Here, a =3, d= 15-3 = 12

Since a_{n }= a_{k }= (n-k) d

a_{n }– a_{54 }= (n-54) 12

132 = 12n – 54 * 12 …..(since a_{n }– a _{54 }= 132 given)

12 n = 132 + 54 * 12 = 12 (11+ 54)

n= 11 + 54 = 65

Find the sum of 2+3+5+9+8+15+11+21+…(2n+1) th term

Let there be n problems as that was fed to the comuter.

Let t be the time it had taken to solve the first question.

So now aaccording to the problem the time taken to solve the successive questions will be t-t/x, t-2t/x,….., t-(n-1)t/x.

There are three variables in the problem – n,t,x.

You need to solve for n.

There are three conditions in the problem. So now use the three conditions and form equations.

Then solve for n.

t-t/x + t-2t/x + t-3t/x + ……. t-(n-1)t/x = 63.5 ———– (1)

t + t-t/x + t-2t/x +…… + t-(n-2)t/x = 127 ———- (2)

t-2t/x + t-3t/x + …….. + t-(n-1)t/x = 31.5 ——— (3)

From the three equations, solve for n.

(1)-(3):

t-t/x = 32 ————- (4)

(2)-(1):

(n-1)t/x = 63.5 ————(5)

From (2):

(n-1)t – (t/x)(n-2)(n-1)/2 = 127

or (n-1)t*[1 – (n-2)/(2x)] = 127

63.5x*[1 – (n-2)/(2x)] = 127 ——– from (5) (n-1)t = 63.5x

or x[1 – (n-2)/(2x)] = 2

or [2x-n-2] = 4

or x-1 = (n+4)/2 ———– (6)

Using (4):

(1-1/x)[63.5x/(n-1)] = 32

(x-1)[63.5/(n-1)] = 32 ——– (7)

So 63.5/(n-1) [n+4] = 64

or 63.5(n+4) = 64(n-1)

So 63.5n + 254 = 64n – 64

or 0.5n = 318.

or n = 636.

So 636 Questions would be fed to the computer.

That solves this question.

Hope it helps.

@Himanshu

if the sum of the 8th and 9th terms of an arithmetic progression is 72 and the 4th term is -6, find the common difference.

Ans is 84/9 or 9.33

first term a And 9th term a+8d sum of these terms is 2a+8d=24 ie a+4d=12 sum of the first 9 numbers is 9/2(2a+8d) ie 9/2(2(a+4d))ie 9/2(2.12)ie 108

@RAHUL

first n odd numbers are

1,3,5……….n

Sn=1+3+5……+n

a=1, d=2

Sn=n/2(2a+(n-1)d)

Sn=n/2(2*1+(n-1)2)

Sn=n/2(2+2n-2)

Sn=n/2(2n)

Sn=n*n

Sn=n square

@qwerty

100 and 200 are divisible of 5

x1=100

xn=200

d=5

n=(xn-x1/d)+1 = (200-100/5)+1 =(100/5)+1=20+1=21

sum= (1/2)n(xn+x1)=21/2(300)=(21)150=3150

@myza sahira

2) xn=52

x1=12

d=5

n=(xn-x1/d)+1

(52-12/5)+1 = (40/5)+1 = 8+1=9

i will find the answer of the first question soon

@Joshua

sum= 20800

sum=1/2n(x1+xn)

20800=1/2(25)(x1+xn)

x1+xn=416

can i have some more problems regarding ap

please help me me with this one, i’m new….

the sum of three numbers in an AP is 21 and their product is 280. find the numbers.

Let the nos be a-d , a, a+d

a-d+a+a+d=21

3a =21

a=7

a-d×a×a+d=280

a²-d²×a=280

49-d²×7=280

49-d²=40

d²=9

d=3

The nos are

6,7and8

@RAHUL

d=2

a=1

sn=n/2(2a+(n-1)d)

sn=n/2(2*1+(n-1)2)

sn=n/2(2+2n-2)

sn=n/2(2n)

sn=n*n

sn=n square

@Abhilash

2. 2140

pls give the procedure to solve the problem

@nikhil

@nikhil

For the 4th question the answer can be 43.

i.e, an-a54=(n-54)12

132=12n-648

12n=132+648

12n=516

n=43.

Why can’t we do in this format?

@myza sahira

n=11

@SOMANATHAN

T-40=252

FIND THE SUM OF FIRST n ODD NATURAL NUMBER.

@SOMENDRA MISHRA

@azimah

The number of terms in an A.P. is even. The sums of odd and even numbered terms are 24 and 30 respectively. If the last term exceeds the first term by 10.t, the number of terms in the A.P. is

the maximum speed of an engine is 1200 p.s.m. and minimum speed is 40 p.s.m. the machine should be equipped with a 5-speed. find 3 more speed needed by using the geometric progression.

please help me solve this problem,.tq.,

Please help me with this one

Find the sum of the A.P. -7-3+1+… from the seventh to the thirtieth term inclusive.

this examples helped me alot

nth term of A.P. is

a + (n-1) d

a is the first term{a1}

d is the common difference{ a2-a1 = a3-a2 = …… = (an) – (an-1) }

The sum of first numbers of ap is 20 and there square is 120 find the numbers

Find the 40th term of an arithematic progression whose 9th term is 465 and 20th term is 388.

send your answers with formula to padmaja_somanathan@yahoo.co.uk

The som ofn terms of two AP are in the ration (2n+3):(6n+5),then the ration of thier 13 th terms is

(a) 53:155 (b)29:83 (c) 31:89 (d)27:87

@myza sahira

Here in question as the AP of the series

2,7,12…52

so we use formula is =

nth = a+(n-1)d

here nth is = 52, a= 2, d = 5

52= 2+(n-1)5= 52-2= 5n-5

= 50 = 5n-5 =50+5 = 5n

= 55= 5n

= n= 55/5

hence n= 11th term

and we also check this

the following formula

nth term = (a+10th d)

11th = 2+ 10*5

= 2+50 = 52 is the answer

@Abhilash

You can use the formula to calculate the AP term.

In this question given that series 12,32,42……and so on and find the 20th terms of the series

formula is that 20th =(a+19d)

here a = 12, d = 10 that is common difference

so 20th = (12+19*10)

= 20th = 12+190 = 202 is the 20th term. ANS .

sum of 4 term of an A.P is an 32 .The ratio of product first and fourth is to be the product of second and third term is 55:63.Find the number?

if 10th term is 17 then find out sum upto 10th term??

In an arithmetic progression, the sum of the first three terms is 18, and the sum of squares of these terms is 126. find the terms.

sum of even natural numbers from 20-100??????????

find the sum of arithmetic series which 25 terms and its middle number is 20???

Sñ=n (a+Tñ/2)

S25=25×20

S25=500

Numbers between 100 and 200 divisible by 5 will be 105, 110, 115,…..195.

Hence apply formula, nth term = Tn=a+(n-1)d

Here, Tn=195, a=105, d=5, which makes n=19.

So, sum of all the terms Sn = n/2(a+l) (l= last term = 195; a = first term=105) which gives 2850.

@qwerty

@myza sahira

Qn. 1 – check for typo error.

Ans for Qn 2. You know the n th term of the AP as 52 and common difference d = 5

Hence, apply the formula Tn = a+(n-1)d

i.e. 52 = 2+(n-1)5, solving this equation for n gives n=11. Hence, there are 11 terms in the series

@Joshua

n=25; d= 35;

sum of n terms of an AP Sn =n/2{2a + (n-1)d}

Here Sn = S25 = 20800= 25/2{2a+(25-1)35}

Solving this equation gives a= 412 (first month arrear)

Hope this has helped.

Last month arrear Sn i.e. S25= a+(n-1)d = 412+24*35 = 1252

the interior angles of convex polygon are in ap, the common difference being 5 degree. if smallest angle is 120 degree then the no of sides is——–

2140

@Abhilash

The sum of first n terms of two A.P are in the ration 3n+8 :70+15. Find the ration of their 3rd term

find the four number of A.P. such that their sum is 38 and their product is 2856

if sum of 13 = 21 and sum of 21=13 prove that sum of 34=-34

Hi friends,

p,q,r are in A.P.prove that

p(q+r)+q(p+r)+r(q+p)=2/9(p+q+r)cube

there are 37 terms in an ap. the sum of the middle three terms is 289 and the product of the last three terms is 4037.find the ap

sum of the nos between 100 and 200 divisible by 5 .thanks

hai..

i want to ask if we want to find the next term in the progression and the questoin is

1),-q,0,p+q… how?

2)find the number of term for the following AP

2,7,12,…,52

please give me the solution..

thankz..

A company owner paid a sum of 20800$ to an employee as a salary arrears in 25 regular installments such that each month’s payment is 35$ more than the preceeding month. Find the arrear amount of the 1st month and the last month.

it was helpful

1) 108

2) 2140 @Abhilash

the sum of the first 20 terms in an AP is 210.

find the sum of the 10th and the 11th terms..

a.)10.5

b.)42

c.)21

d.)can not be determined

Each year a tree grows 5 cm less than it did in the preceding year. If it grew 1 cm in first year in how many years it have ceased growing?

A computer solve several problems in succession. The time it took the computer to solve each successive problem was the same number of times smaller than the time it took to solve the preceding problem. How many problems were suggested to the computer if it spent 63.5 min to solve all the problems except for the first,127 min to solve all the problems except for the last one,and 31.5 min to solve all the problems except for the first two?

1. The sum of the first and the 9th of an arithmetic progression is 24. What is the sum of the first nine

terms of the progression ?

2. The sum of 20 terms of the series 12 + 22, 32 + 42, 52 + 62 is?