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Byju's Answer

Standard VIII

Mathematics

Division of an Expression by a Expression

The sum of fi...

Question

The sum of first $n$ terms of an infinite G.P. is

S = \frac{a}{1 - r}

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S_{n} = \frac{a_{1} (1 - r^{n})}{1 - r}

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S = \frac{a n}{1 - r}

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S_{n} = \frac{a_{1} (1 - r^{n})}{1 + r}

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Solution

The correct option is A $S = \frac{a}{1 - r}$
Sum of GP $= \frac{a (r^{n} - 1)}{r - 1}$
$a =$ first term
$r =$ common ratio
For $n \to \infty$
$r^{n} = 0$ for $r < 1$
$r^{n} \to \infty$ for $r > 1$
Thus $sum = \frac{a}{r - 1}$

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Similar questions

Q. If

S_{n}

represents the sum of

n

terms of a

G . P .

whose first term and common ratio are

a

and

r

respectively, then prove that

S_{1} + S_{2} + S_{3} + . . . + S_{n} = \frac{n a}{1 - r} - \frac{a r (1 - r^{n})}{{(1 - r)}^{2}}

Q. Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. prove that

P^{2} R^{n} = S^{n}

Q. For first

n

natural numbers we have the following results with usual notations

n \sum r = 1 r = \frac{n (n + 1)}{2}, n \sum r = 1 r^{2} = \frac{n (n + 1) (2 n + 1)}{6}, n \sum r = 1 r^{3} = {(n \sum r = 1 r)}^{2}

a_{1} a_{2} . . . . a_{n} \in A . P

then sum to

n

terms of the sequence

\frac{1}{a_{1} a_{2}}, \frac{1}{a_{2} a_{3}}, . . . \frac{1}{a_{n - 1} a_{n}}

is equal to

\frac{n - 1}{a_{1} a_{n}}

and the sum to

n

terms of a

G . P

with first term '

a

' & common ratio '

r

' is given by

S_{n} = \frac{l r - a}{r - 1}

for

r \neq 1

for

r = 1

sum to

n

terms of same

G . P .

n

a

, where the sum to infinite terms of

G . P .

is the limiting value of

\frac{l r - a}{r - 1}

when

n \to \infty, | r | < l

where

l

is the last term of

G . P .

On the basis of above data answer the following questionsThe sum to infinite terms of the series

\frac{1}{2} + \frac{1}{6} + \frac{1}{18} + . .

is equal to ?

Q. If

S_{n}

represents the sum of

n

terms of a

G . P .

whose first term and common ratio are

a

and

r

respectively, then prove that

S_{1} + S_{3} + S_{5} + . . . + S_{2 n - 1} = \frac{a n}{1 - r} - \frac{a r (1 - r^{2 n})}{{(1 - r)}^{2} (1 + r)}

Q. The set

{a_{1}, a_{2}, . . . . a_{n}}

form a

G . P .

with first term as

a

and common ratio

r

. Prove that

\sum a_{i} a_{j} = \frac{a^{2} r ({1 - r}^{n - 1}) ({1 - r}^{n})}{{(1 - r)}^{2} (1 + r)}

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The sum of first n terms of an infinite G.P. is

The correct option is A S=a1−rSum of GP =a(rn−1)r−1a= first termr= common ratioFor n→∞rn=0 for r<1rn→∞ for r>1Thus sum=ar−1

The sum of first $n$ terms of an infinite G.P. is

The correct option is A $S = \frac{a}{1 - r}$
Sum of GP $= \frac{a (r^{n} - 1)}{r - 1}$
$a =$ first term
$r =$ common ratio
For $n \to \infty$
$r^{n} = 0$ for $r < 1$
$r^{n} \to \infty$ for $r > 1$
Thus $sum = \frac{a}{r - 1}$