Let a_1,a_2,,a_30 be an A.P., S=_i=1^30a_i and T=_i=1^15a_(2i-1).

Question

Let $a_1,a_2,\ldots,a_{30}$ be an A.P., $S=\displaystyle\sum_{i=1}^{30}a_i$ and $T=\displaystyle\sum_{i=1}^{15}a_{(2i-1)}$. If $a_5=27$ and $S-2T=75$, then $a_{10}$ is equal to:

Accepted Answer

Let the common difference = d

S = $$\sum\limits_{i = 1}^{30} {{a_i}} $$

=  $a$1 + $a$2 + . . . . . + $a$30

$$ \therefore $$  S = $${{30} \over 2}\left[ {{a_1} + {a_{30}}} \right]$$

= 15 [$a$1 + $a$1 + 29d]

=  15 (2$a$1 + 29d)

T  =  $$\sum\limits_{i = 1}^{15} {{a_{\left( {2i - 1} \right)}}} $$

=  $a$1 + $a$3 + . . . . . . + $a$29

=  $${{15} \over 2}\left[ {a{}_1 + {a_{29}}} \right]$$

=  $${{15} \over 2}\left[ {a{}_1 + {a_1} + 28d} \right]$$

=  $${{15} \over 2}\left[ {2a{}_1 + 28d} \right]$$

=  15 ($a$1  + 14d)

Given,

S - 2T = 75

$$ \Rightarrow $$  15(2$a$1 + 29d) - 2 $$ \times $$ 15 ($a$1 + 14d) = 75

$$ \Rightarrow $$  30$a$1 + 15 $$ \times $$ 29d - 30 $a$1  - 420d = 75

$$ \Rightarrow $$  435d - 420d = 75

$$ \Rightarrow $$  15d = 75

$$ \Rightarrow $$  d = 5

Given that,

$a$5  = 27

$$ \Rightarrow $$  $a$1  + 4d  = 27

$$ \Rightarrow $$  $a$1 + 20 = 27

$$ \Rightarrow $$  $a$1 = 7

$$ \therefore $$  $a$10  = $a$1 + 9d

= 7 + 45

=  52

Solutions · 1