HOW TO FIND A AND D OF AN ARITHMETIC SEQUENCE GIVEN THE SUM

Question 1 :

Given  :  a  = 3 , n = 8 ,S n = 192 find d

Solution :

S n  =  (n/2) [2a + (n - 1) d]

 192  =  (8/2) [2 (3) + (8 - 1) d]

 192  =  4 [6 + 7 d]

 192/4  =  [6 + 7 d]

  48 - 6  =  7 d

    42/7  =  d

      d  =  6

Question 2 :

Given l = 28 , S = 144 , and there are total 9 terms. Find a

Solution :

S n = (n/2) [a + l]

n  =  9

144  =  (9/2) [ a + 28]

(144 x 2)/9  =  a + 28

  32  =  a + 28

  a  =  32 - 28

   a  =  4

Question 2 :

How many terms of the AP 9, 17, 25,.......... must be taken to give a sum of 636?

Solution :

a_n = a + (n - 1) d

S_n = 636

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

a = 7   d = 17 - 9 = 8, Sn = 636      

636  =  (n/2)[2(9) + (n-1) 8]

(636 x 2)  =  n[18 + 8 n - 8]

(636 x 2)  =  n[10 + 8 n]

636  =  n[5 + 4n]

636  =  5n + 4 n²

4 n² + 5 n - 636  =  0

(4n + 53)(n - 12)  =  0

 n = -53/4         n = 12

Hence 12 terms are need to give a sum of 636.

Question 3 :

The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and common difference.

Solution :

 a = 5, l = 45, S_n = 400

 S_n  =  (n/2)[a + l]

 400  =  (n/2)[5 + 45]

  400  =  (n/2)[50]

   400  =  n(25)

  400/25  =  n

  n  =  16

 Sn  =  (n/2)[2 a + (n- 1) d]

400  =  (16/2) [2(5) + (16-1)d]

400  =  8 [10 + 15 d]

  400/8  =  10 + 15 d

50  =  10 + 15 d

 50 - 10  =  15 d

40  =  15 d

  40/15  =  d

   d  =  8/3

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.