QJEEM19Ma4af6: JEE Main 2019 Mathematics - Class 11 JEEM19Ma4af6 Solution: Let be an A.P., and . If = 27 and S - 2T = 75, then is equal to :

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