Sum of first and third term of an arithmetic progression is 12 and product of first and second term

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8. Sequences and Series

easy

The sum of the first and third term of an arithmetic progression is $12$ and the product of first and second term is $24$, then first term is

A

$1$

B

$8$

C

$4$

D

$6$

Solution

(c) Let first $3$ terms be $a – d,a\,\,$and $a + d$

Now $(a – d) + (a + d) = 12$

$ \Rightarrow $$2a = 12$

$ \Rightarrow $ $a = 6$ and $(a – d)a = 24$

$⇒$ $6(6 – d) = 24$

$⇒$ $d = 2$

First term = $a – d = 6 – 2 = 4$.

Standard 11

Mathematics

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