Let be an arithmetic progression. If (where ) then

Question

Let $a_{1}, a_{2}, a_{3}, \dots$ be an arithmetic progression. If $\frac{a_{1} + a_{2} + \dots + a_{10}}{a_{1} + a_{2} + \dots + a_{P}} = \frac{100}{P^{2}}$ (where $P \neq 0$ ) then $\frac{a_{11}}{a_{10}} =$

Smart Achievers · Accepted Answer

(3)

$\frac{S_{10}}{S_{p}} = \frac{100}{p^{2}} \Rightarrow S_{p} = \frac{S_{10} p^{2}}{100}, \frac{a_{11}}{a_{10}} = \frac{S_{11} - S_{10}}{S_{10} - S_{9}} = \frac{21}{19}$

Solution

Q.
Let $a_{1}, a_{2}, a_{3}, \dots$ be an arithmetic progression. If $\frac{a_{1} + a_{2} + \dots + a_{10}}{a_{1} + a_{2} + \dots + a_{P}} = \frac{100}{P^{2}}$ (where $P \neq 0$ ) then $\frac{a_{11}}{a_{10}} =$

1 $\frac{19}{21}$

2 $\frac{100}{21}$

3 $\frac{21}{19}$

4 $\frac{121}{100}$

Ans.
(3)

$\frac{S_{10}}{S_{p}} = \frac{100}{p^{2}} \Rightarrow S_{p} = \frac{S_{10} p^{2}}{100}, \frac{a_{11}}{a_{10}} = \frac{S_{11} - S_{10}}{S_{10} - S_{9}} = \frac{21}{19}$

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Solution

Q. Let a1,a2,a3,… be an arithmetic progression. If a1+a2+⋯+a10a1+a2+⋯+aP=100P2 (where P≠0) then a11a10=

1 1921

2 10021

3 2119

4 121100

Ans. (3) S10Sp=100p2⇒Sp=S10p2100, a11a10=S11-S10S10-S9=2119