If $a_{1}$ , $a_{2}$ , $a_{3}$ , ... are in AP then $a_{p}$ , $a_{q}$ , $a_{r}$ are in AP if $p, q, r$ are in

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Solution

The correct option is A AP
$a_{p}, a_{q}, a_{r}$ are in AP

$\Rightarrow 2 a_{q} = a_{p} + a_{r}$
$\Rightarrow 2 a_{1} + 2 (q - 1) d = a_{1} + (p - 1) d + a_{1} + (r - 1) d$
$\Rightarrow 2 q d - 2 d = p d - d + r d - d$
$\Rightarrow 2 q = p + r$ (for $d \neq 0$ )
$\Rightarrow p, q, r$ are in AP
$∴$ Ans. is option A.

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If a1, a2, a3, ... are in AP then ap, aq, ar are in AP if p,q,r are in

The correct option is A APap,aq,ar are in AP⇒2aq=ap+ar⇒2a1+2(q−1)d=a1+(p−1)d+a1+(r−1)d⇒2qd−2d=pd−d+rd−d⇒2q=p+r (for d≠0) ⇒p,q,r are in AP ∴ Ans. is option A.

If $a_{1}$ , $a_{2}$ , $a_{3}$ , ... are in AP then $a_{p}$ , $a_{q}$ , $a_{r}$ are in AP if $p, q, r$ are in