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Formula for Sum of n Terms of an AP

The sucessive...

Question

The sucessive terms of an $A P$ are $a_{1}, a_{2}, a_{3}, . . . .$ If $a_{6} + a_{9} + a_{12} + a_{15} = 20$ then $\sum_{r = 1}^{20} a_{r}$ is equal to

A

75

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B

100

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C

120

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D

150

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Solution

The correct option is B $100$
or, $a + 5 d + a + 8 d + a + 11 d + a + 14 d = 20$
or, $4 a + 38 d = 20$
or, $2 a + 19 d = 10 - - - (1)$
Now, $20 \sum r = 1 a_{r} = S_{20} = \frac{20}{2} [a_{1} + a_{20}]$
$= 10 [a + a + 19 d]$
$= 10 [2 a + 19 d]$ [from $(1)$ ]
$= 10 \times 10 = 100$

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Q. The successive terms of an A.P. are

a_{1}, a_{2}, a_{3}, \dots \dots . I f a_{6} + a_{9} + a_{12} + a_{15} = 20, t h e n \sum_{r = 1}^{20} a r =

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

be terms of an A.P. If

\frac{a_{1} + a_{2} + \dots + a_{p}}{a_{1} + a_{2} + \dots + a_{q}} = \frac{p^{2}}{q^{2}} (p \neq q)

\frac{a_{6}}{a_{21}}

a_{1}, a_{2}, a_{3}, . . . .

be terms of an A.P. If

\frac{a_{1} + a_{2} + . . . . . + a_{p}}{a_{1} + a_{2} + . . . . . + a_{q}} = \frac{p^{2}}{q^{2}}, p \neq q

\frac{a_{6}}{a_{21}}

a_{1}, a_{2}, a_{3} . . . .

be terms of an A.P. If

\frac{a_{1} + a_{2} + . . . . + a_{p}}{a_{1} + a_{2} + . . . . + a_{q}} = \frac{p^{2}}{q^{2}}, p \neq q

\frac{a_{6}}{a_{21}}

Let $a_{1}, a_{2}, a_{3}, . . .$ be the terms of an AP. If $\frac{a_{1} + a_{2} + a_{3} + . . . + a_{p}}{a_{1} + a_{2} + a_{3} + . . . + a_{q}} = \frac{p^{2}}{q^{2}}, p \neq q,$ find the value of $\frac{a_{6}}{a_{21}}$ .

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