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Sum of n Terms

Let a1, a2,...

Question

Let $a_{1}, a_{2}, a_{3}, . . . . . . . . . . a n d b_{1} b_{2}, b_{3} . . . . . . . . . . .$ be arithmetic progression such that $a_{1} = 25, b_{1} = 75 a n d a_{100} + b_{100} = 100$ , then

A

The common difference in progression

^{'} a_{i}^{'}

is equal but opposite in sign to the common difference in progression

^{'} b_{j}^{'} .

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B

a_{n} + b_{n}

= 100 for any n.

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C

(a_{1} + b_{1}), (a_{2} + b_{2}), (a_{3} + b_{3}),

.......... are in A.P.

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D

100 \sum r = 1 (a_{r} + b_{r}) = 10^{4}

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Solution

The correct options are
A The common difference in progression $^{'} a_{i}^{'}$ is equal but opposite in sign to the common difference in progression $^{'} b_{j}^{'} .$
B $a_{n} + b_{n}$ = 100 for any n.
C $100 \sum r = 1 (a_{r} + b_{r}) = 10^{4}$
D $(a_{1} + b_{1}), (a_{2} + b_{2}), (a_{3} + b_{3}),$ .......... are in A.P.
Given $a_{1}, a_{2}, a_{3}, . . . . . . . . . .$ and $b_{1} b_{2}, b_{3} . . . . . . . . . . .$ be two AP's.
Let $d_{1}, d_{2}$ be the common difference of the two AP's.
Then, the AP's can be written as
$a_{1}, a_{1} + d_{1}, a_{1} + 2 d_{1} . . . . . . . . . .$
$b_{1}, b_{1} + d_{2}, b_{1} + 2 d_{2} . . . . . . . . . .$
Now, given $a_{100} + b_{100} = 100$
$\Rightarrow a_{1} + 99 d_{1} + b_{1} + 99 d_{2} = 100$
$\Rightarrow 100 + 99 (d_{1} + d_{2}) = 100$
$\Rightarrow d_{1} + d_{2} = 0$
$\Rightarrow d_{1} = - d_{2}$
Hence, option A is correct.
Now, $a_{n} + b_{n}$
$= a_{n} + (n - 1) d_{1} + b_{1} + (n - 1) d_{2}$
$= a_{1} + b_{1} = 100$
Hence, option B is correct.
Since, $a_{n} + b_{n} = 100 \forall n$
So, $a_{1} + b_{1}, a_{2} + b_{2} + . . . .$ are in AP.
Hence, option C is also true.
Now $100 \sum r = 1 (a_{r} + b_{r}))$
$= a_{1} + b_{1}, a_{2} + b_{2} + . . . . + a_{99} + b_{99} + a_{100} + b_{100}$
$= 100 + 100 + . . . . u p t o 100 t i m e s$
$= 10^{4}$
Hence, option D is also true.

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Similar questions

Q. Given a geometric progression

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

and an arithmetic progression

b, b_{1}, b_{2}, b_{3}, . . . . . . . . .

with positive terms. The common difference of A.P. and common ratio of G.P. are both positive. Show that there always exists a system of logarithms for which

l o g a_{n} - b_{n} = l o g a - b

(for any n)
Find base

β

b_{i} > 1 for i = 1, 2, . . ., 101.

{log}_{e} b_{1}, {log}_{e} b_{2}, . . ., {log}_{e} b_{101}

are in Arithmetic Progression (A.P.) with the common difference

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a_{1}, a_{2}, . . ., a_{101}

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a_{51} = b_{51} .

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{log}_{e} b_{1} {log}_{e} b_{2}, . . ., {log}_{e} b_{101}

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a_{1}, a_{2}, . . . a_{101}

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a_{51} = b_{51}

t = b_{1} + b_{2} + . . . . + b_{51}

s = a_{1} + a_{2} + . . . . + a_{51}

a_{1}, a_{2}, a_{3}, . . . (a_{i} \neq a_{j} \neq 0)

b_{1}, b_{2}, b_{3}, . . . (b_{i} \neq b_{j} \neq 0)

are two different A.P., then which of the following is/are always correct?

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(a_{1} + b_{1}), (a_{2} + b_{2})

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