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Standard XII

Mathematics

A.G.P

Suppose a ser...

Question

Suppose a series of $n$ terms is given by $S_{n} = a_{1} + a_{2} + . . . + a_{n}$ Then $S_{n - 1} = a_{1} + a_{2} + . . . + a_{n - 1} (\forall n > 1)$ and then we can write $t_{n} = S_{n} - S_{n - 1} \forall n \geq 2$ for $n = 1$ we have $S_{1} = t_{1}$
On the basis of above information answer the following questionsThe sum to $n$ terms of a series is $a {.3}^{n} - b$ where $a$ & $b$ are constants, then the series is

A . P .

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A . G . P

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G P .

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G . P

from second terms onwards

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Solution

The correct option is B $G . P$ from second terms onwards
$S_{n} = a {.3}^{n} - b$
$t_{1} = S_{1} = 3 a - b$
$t_{n} = S_{n} - S_{n - 1} = 2 a 3^{n - 1}$
$∴$ The series is $3 a - b, 6 a, 18 a, 54 a, . . .$
It is a GP from $2^{n d}$ term onwards

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Similar questions

Q. Suppose a series of

n

terms is given by

S_{n} = a_{1} + a_{2} + . . . + a_{n}

Then

S_{n - 1} = a_{1} + a_{2} + . . . + a_{n - 1} (\forall n > 1)

and then we can write

t_{n} = S_{n} - S_{n - 1} \forall n \geq 2

for

n = 1

we have

S_{1} = t_{1}

On the basis of above information answer the following questionsThe sum to

n

terms of a series is

\frac{n (n + 1) (n + 2)}{3}

then its

n^{t h}

term is given by

Q. Assertion :If

S_{n}

denotes the sum of

n

terms a series given by

S_{n} = \frac{n (n + 1) (n + 2)}{6} \forall n \geq 1,

then

lim n \to \infty n \sum r = 1 \frac{1}{t_{r}} = 4

Reason:

t_{n} = S_{n} - S_{n - 1}

Q. Consider a series whose sum to n-terms is denoted by

S_{n} & n^{t h}

term is

t_{n}

where

S_{n} = n^{3} + 2^{n + 1} \forall n \in N

, then the value of

(t_{5} - t_{3})

is equal to

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 of the series

\frac{1}{2.3} + \frac{1}{3.4} + . . .

n

terms is?

Q. If

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

are in A.P. with

s_{n}

as the sum of first 'n' terms

(S_{0} = 0), t h e n \sum_{k = 0}^{n}^{n} C_{k} S_{k}

is equal to

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The correct option is B G.P from second terms onwardsSn=a.3n−bt1=S1=3a−btn=Sn−Sn−1=2a3n−1 ∴ The series is 3a−b,6a,18a,54a,...It is a GP from 2nd term onwards

The correct option is B $G . P$ from second terms onwards
$S_{n} = a {.3}^{n} - b$
$t_{1} = S_{1} = 3 a - b$
$t_{n} = S_{n} - S_{n - 1} = 2 a 3^{n - 1}$
$∴$ The series is $3 a - b, 6 a, 18 a, 54 a, . . .$
It is a GP from $2^{n d}$ term onwards