Question

If the 8th term of an AP is 31 and its 15th term is 16 more than the 11 the term, find AP.
Or, find the sum of all two-digit odd positive numbers.

Answer 1

Here,
a₈ = 31 and a₁₅ = 16 + a₁₁
⇒ 31 = a + 7d .............(i)
and 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 two - digit odd positive numbers are 11, 13, 15, 17 ......99.
This is an AP in which a = 11, d = (13 - 11) = 2 and l = 99.
Let the number of terms be n. Then,
T_n = 99
⇒ a + (n - 1) × d = 99
⇒ 99 = 11 + (n - 1) × 2
⇒ 99 = 11 + 2n - 2
⇒ 90 = 2n
⇒ n = 45
Required sum = $\frac{n}{2}\left(a+l\right)$
= $\frac{45}{2}\left(11+99\right)=\left(\frac{45}{2}\times 110\right)$
= (45 × 55) = 2475