If 7 = 5 + 7 1 ( 5 + α ) + 7 2 1 ( 5 + 2 α ) + 7 3 1 ( 5 + 3 α ) + … ∞ , then the value of α is :

S = a + ( a + d ) r + ( a + 2 d ) r 2 + … Then S = 1 − r a + ( 1 − r ) 2 d r , ∣ r ∣ < 1 Since, r = 7 1 and a = 5 , d = α 7 = 1 − 7 1 5 + ( 1 − 7 1 ) 2 α ⋅ 7 1 ⇒ α = 6

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MathematicsSequence & SeriesA.P.Medium2 minPYQ_2025

MathematicsMediumsingle choice

If $7 = 5 + \frac{1}{7} (5 + α) + \frac{1}{7 ^{2}} (5 + 2 α) + \frac{1}{7 ^{3}} (5 + 3 α) + \dots \infty$ , then the value of $α$ is :

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Answer:

Solution:

$S = a + (a + d) r + (a + 2 d) r^{2} + \dots$
Then $S = \frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}}, ∣ r ∣ < 1$
Since, $r = \frac{1}{7}$ and $a = 5, d = α$
$7 = \frac{5}{1 - \frac{1}{7}} + \frac{α \cdot \frac{1}{7}}{( 1 - \frac{1}{7} ) ^{2}} \Rightarrow α = 6$

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:A.P.

⏱ 2mℹ️ Source: PYQ_2025

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