Question

An arithmetic sequence has $6^{th}$ term as 52 and ${15}^{th}$ term as 142. Find a & d .

• A. 2,10
• B. 10,2
• C. 2,20
• D. 20,2

Answer 1

The correct option is A

2,10

given $a_6$ = a + ( 6 - 1 ) d = 52

a + 5d = 52 -------------------------------------------------------------------(i)

and $a_{15}$ = a + ( 15 - 1 ) d = 142

a + 14d = 142 ----------------------------------------------------------------(ii)

Solving (i) and (ii)

(ii) – (i) $\Rightarrow$ 9d = 90

d = 10

Substituting d=10 in (i)

a+50 = 52

a = 2