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Sum of n Terms

If a 1 , a ...

Question

If $a_{1}, a_{2}, a_{3}, a_{4}, . . {. ., a}_{20}$ are in AP and $a_{1} + a_{20} = 45$ , then $a_{1} + a_{2} + a_{3} + a_{4}, . . {. . + a}_{20}$ is equal to

A

90

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B

900

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C

350

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D

450

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E

730

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Solution

The correct option is E $450$
$∵$ Given $a_{1}, a_{2}, a_{3}, . . . . a_{20}$ are in AP and
$a_{1} + a_{20} = 45.. (i)$
sum of first $n$ terms $= \frac{n}{2} (a + l)$
$a + l = 45$
$[∵ a_{1} + a_{20} = 45.]$
Have, $n = 20$
$∴ a_{1} + a_{2} + a_{3} + . . . . {+ a}_{20} = \frac{20}{2} \times 45 = 450$

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a_{1}, a_{2}, a_{3}, a_{4}

are in AP, then

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