Out of four number in a sequence,first three are in GP with first term 2 and last three are in AP with common difference 12. Find the common ratio, if the number are positive.

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Solution

The correct option is B
3

$t_{1}$ = 2r

$t_{2}$ = 2r

$t_{3}$ = 2 $r^{2}$

There are the first three terms which are in GP last three terms in AP are

$T_{2}$ = 2

$T_{3}$ = 2r + 12

$T_{4}$ = 2r + 24

Equating $t_{3}$ in AP and GP

2r + 12 = 2 $r^{2}$

$r^{2}$ - r - 6 = 0

(r - 3) (r + 2) = 0 $\Rightarrow$ r = 3 [r is +ive]

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Out of four number in a sequence,first three are in GP with first term 2 and last three are in AP with common difference 12. Find the common ratio, if the number are positive.

The correct option is B 3 t1 = 2r t2 = 2r t3 = 2r2 There are the first three terms which are in GP last three terms in AP are T2 = 2 T3 = 2r + 12 T4 = 2r + 24 Equating t3 in AP and GP 2r + 12 = 2r2 r2 - r - 6 = 0 (r - 3) (r + 2) = 0 ⇒ r = 3 [r is +ive]