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nth Term of A.P

Let an be a...

Question

Let $a_{n}$ be an A.P. for which $a_{2} = 20$ and $a_{10} = 40$ . Find $a_{5}$

A

0

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B

10

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C

20

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D

30

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Solution

The correct option is D 30
Applying Arithmetic mean formula,
$a = \frac{2 + 10}{2} = a_{5} = \frac{20 + 40}{2}$
$a_{5} = \frac{60}{2}$
$a_{5} = 30$

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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}, . . . . 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}

If for an A.P. $a_{5} = a_{10} = 5 \times a$ , then $a_{15}$ is....

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}}

Q. Write the expression

a_{n} - a_{k}

a, a + d, a + 2 d, . . . . .

Hence, find the common difference of the A.P for which

a_{10} - a_{5} = 200

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