Given:
5th term of an AP = 31
9th term = 59
Find:
the sum of the first 14 terms of an arithmetic sequence
Solution:
We know that,
nth term of an AP – a(n) = a + (n - 1)d
Hence,
⟹ a + (5 - 1)d = 31
⟹ a + 4d = 31 -- equation (1)
Similarly,
⟹ a + (9 - 1)d = 59
⟹ a + 8d = 59 -- equation (2)
Subtract equation (1) from (2).
⟹ a + 8d - ( a + 4d ) = 59 - 31
⟹ a + 8d - a - 4d = 28
⟹ 4d = 28
⟹ d = 28/4
⟹ d = 7
Substitute the value of d in equation (1).
⟹ a + 4(7) = 31
⟹ a + 28 = 31
⟹ a = 31 - 28
⟹ a = 3
Now,
Sum of first n terms of an AP – S(n) = n/2 * [ 2a + (n - 1)d ]
⟹ S(14) = 14/2 * [ 2(3) + (14 - 1)(7) ]
⟹ S(14) = 7 [ 6 + 91 ]
⟹ S(14) = 7 * 97
⟹ S(14) = 679
∴ The sum of first 14 terms of the given AP is 679.
I hope it will help you.
Regards.