The sum of the first 7 terms of an AP is 63 and that of its next 7 terms is 161.
Find the AP.
Answers
Sn = ( n / 2) [ 2a + ( n -1)d
sum of the first 7 terms of an A.P is 63 i. e S7 = 63.
( 7 / 2) [ 2a + 6d ] = 63
2a + 6d = 18 --------(1)
Sum of its next 7 terms = 161.
Sum of first 14 terms = sum of first 7 terms + sum of next 7 terms.
S14 = 63 + 161 = 224
( 14 / 2) [ 2a + 13d ] = 224.
7 [ 2a + 13d ] = 224.
⇒ [ 2a + 13d ] = 32 -------92)
By Solving equation (1) and (2) we obtain
d = 2
a = 3.
therefore AP= 3,5,9,11,13......
Answer:
3, 5, 7, 9 ....
Step-by-step explanation:
Using S = (n/2) [2a + (n - 1)d] ,where letters have their usual meaning.
For first 7 terms:
⇒ 63 = (7/2) [2a + (7 - 1)d]
⇒ (63 * 2)/7 = [2a + 6d]
⇒ 18 = 2a + 6d
⇒ 9 = a + 3d ⇒ 9 - 3d = a
For next 7 terms:
⇒ 161 = (7/2) [8th term + 14th term]
⇒ (161 x 2)/7 = [a + 7d + a + 13d]
⇒ 46 = 2a + 20d
⇒ 23 = a + 10d
⇒ 23 = 9 - 3d + 10d [from above]
⇒ 2 = d
thus, a = 9 - 3(2) = 3
∴ AP is a , a + d, a + 2d...
3, 3 + 2, 3 + 2(2)...
3 , 5 , 7 ....