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Sum of n Terms

Let a1, a2,...

Question

Let $a_{1}, a_{2}, . . . . a_{30}$ be an A.P., $S = 30 \sum i = 1 a_{i}$ and $T = 15 \sum i = 1 a_{(2 i - 1)}$ . If $a_{5} = 27$ and $S - 2 T = 75$ , then $a_{10}$ is equal to

A

57

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B

53

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C

56

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D

52

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Solution

The correct option is D $52$
Let the common difference be $d$ then
$a_{1} + a_{2} + a_{3} + . . . . . + a_{30} = S = \frac{30}{2} [2 a_{1} + 29. d]$
Also $a_{1} + a_{3} + a_{5} + a_{7} + . . . . + a_{29} = T = \frac{15}{2} [2 a_{1} + 28 d]$
Now $A_{5} = 27 \Rightarrow a + 4 d = 27 . . . . (1)$
Also $S - 2 T = 75 = 15 [2 a_{1} + 29. d] - 15 [2 a_{1} + 28 d]$
$\Rightarrow 15. d = 75 \Rightarrow d = 5$
From equation $(1) \Rightarrow a + 20 = 27$
$\Rightarrow a = 7$
$a_{10} = a + 9 d = 7 + 9 (5) = 52 (D)$

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Similar questions

a_{1}, a_{2}, . . . . . ., a_{30}

S = 30 \sum i = 1 a_{i}

T = 15 \sum i = 1 a_{2 i - 1}

a_{5} = 27

S - 2 T = 75

a_{10}

a_{1}, a_{2}, \dots, a_{30}

S = 30 \sum i = 1 a_{i}

T = 15 \sum i = 1 a_{2 i - 1}

a_{5} = 27

S - 2 T = 75

, then the value of

a_{10}

a_{1}, a_{2}, a_{3}, \dots

\frac{a_{1} + a_{2} + \dots + a_{10}}{a_{1} + a_{2} + \dots + a_{p}} = \frac{100}{p^{2}},

p \neq 10,

\frac{a_{11}}{a_{10}}

a_{1}, a_{2}, . . . . ., a_{10}

\frac{a_{3}}{a_{1}} = 25

\frac{a_{9}}{a_{5}}

a_{1}, a_{2}, \dots, a_{10}

\frac{a_{3}}{a_{1}} = 25

\frac{a_{9}}{a_{5}}

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