Question

If the 8th term of an AP is 31 and its 15th term is 16 more than the 11th term, find the AP.
Or, find the sum of all odd integers between 2 and 100 that are divisible by 3.

Answer 1

Here,
a₈ = 31 and a₁₅ = 16 + a₁₁
⇒ 31 = a + 7d .............(i)
And a₁₅ = 16 + a₁₁
⇒ a + 14d = 16 + a + 10d .............(ii)
By solving equation (ii), we get:
4d = 16
⇒ d = 4
On substituting d = 4 in (i), we get:
31 = a + 7 × 4
⇒ a = 3
Since a = 3 and d = 4, required AP = 3, 7 , 11, 15 .....

OR

All odd integers between 2 and 100 divisible by 3 are
3, 9 , 15 , 21 ................ 99
Here a = 3 , d = 6 and n^th term = 99
Let the number of terms be n. Then,
T_n = 99
⇒ a + (n - 1)× d
⇒ 99 = 3+ (n - 1) × 6
⇒ 99 = 3+ 6n - 6
⇒ 102 = 6n
⇒ n = 17
Required sum = $\frac{n}{2}\left(a+l\right)$
= $\frac{17}{2}\times \left(3+99\right)=\left(17\times 51\right)=867$