HOW TO FIND THE SUM OF GIVEN NUMBER OF TERMS FROM INDICATED TERMS

Question 1 :

Find the sum of first 22 terms of an AP in which d = 7 and 22^nd term is 149. 

Solution :

n  =  22, d  =  7

a₂₂  =  a + 21d  =  149  ----(1)

By applying the value of d in (1), we get 

a + 21(7)  =  149

a + 147  =  149

a  =  149 - 147 

a  =  2

Now, we have to find sum of 22 terms.

S_n  =  (n/2) [2a + (n - 1)d]

  =  (22/2) [2(2) + (22 - 1)7]

  =  11 [4 + 21(7)]

  =  11 [4 + 147]

  =  11 (151)

  =  1661

Question 2 :

The sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution :

Given that :

Second term (a₂)  =  14

a + d  =  14 -----(1)

Third term (a₃)  =  18

a + 2d  =  18 -----(2)

(1) - (2) 

(a + d) - (a + 2d)  =  14 - 18

d - 2d  =  -4

-d  =  -4

d  =  4

By applying the value of d in (1), we will get "a".

a + 4  =  14

a  =  14 - 4  =  10

Now, we have to find the sum of 51 terms.

Sn  =  (n/2) [2a + (n - 1)d]

  =  (51/2)[2(10) + (51 - 1)4]

  =  (51/2) [ 20 + 200]

  =  (51/2)(220)

  =  5610

Question 3 :

The 15th term of the AP 13, 17, 21, 25,… is

(a) 102     (b) 104      (c) 69      (d) 108

Solution :

13, 17, 21, 25,…

a = 13, d = 17 - 13 ==> 4 and n = 15

an = a + (n - 1) d

a15 = a + 14d

= 13 + 14(4)

= 13 + 56

= 69

So, option c is correct.

Question 4 :

The sum of first 16th terms of the AP 10, 6, 2, ….is

Solution :

a = 10, d = 6 - 10 ==> -4 and n = 16

S_n = (n/2) [2a + (n - 1)d]

= (16/2) [2(10) + (16 - 1)(-4)]

= 8[20 + 15(-4)]

= 8[20 - 60]

= 8(-40)

= -320

So, the sum of first 16 terms is -320.

Question 5 :

The sum of first three terms of an AP is 33. If the product of first and third term exceeds the second term by 29, find the AP

Solution :

Let the first three terms are a - d, a and a + d.

a - d + a + a + d = 33

3a = 33

a = 33/3

a = 11

(a - d) (a + d) = a + 29

a² - d² = a + 29

Applying the value of a, we get

11² - d² = 11 + 29

121 - d² = 40

 d² = 121 - 40

 d² = 81

d = -9 and 9.

So, the terms are 20, 11, 2 or 2, 11, 20.

Question 6 :

In an AP, if 𝑎 = 1, 𝑎_𝑛 = 20, 𝑆_𝑛 = 399 then n is equal to

a) 19    b) 21    c) 38      d) 42

Solution :

First term = 1

a + (n - 1)d = 20

1 + (n - 1)d = 20

(n - 1)d = 19

𝑆_𝑛 = 399

(n/2)[2a + (n - 1)d] = 399

(n/2)[2(1) + 19] = 399

(n/2)[21] = 399

n/2 = 399/21

n/2 = 19

n = 19(2)

n = 38

So, option c is correct.

Question 7 :

If the 𝑛𝑡ℎ term of an AP is given by 𝑎_𝑛 = 5𝑛 − 3 then the sum of first 10 terms if

a) 225    b) 245   c) 255  d) 270

Solution :

𝑎_𝑛 = 5𝑛 − 3

First term = 2

Common difference = 7 - 2 ==> 5

Sum of 10 terms :

= (n/2)[2a + (n - 1)d]

= (10/2)[2(2) + (10 - 1)(5)]

= 5[4 + 9(5)]

= 5[4 + 45]

= 5(49)

= 245

So, the sum of first 10 terms is 245.

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