If $s_{1} = a_{2} + a_{4} + a_{6} + . . . .$ upto 100 terms and $s_{2} = a_{1} + a_{3} + a_{5} + . . . .$ upto 100 terms of a certain A.P., then its common difference is:

\frac{s_{1} - s_{2}}{100}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

s_{2} - S_{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

s_{1} - S_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{s_{1} - s_{2}}{100}$
Given that,

$S_{1} = a_{2} + a_{4} + - - - - - u p t o 100 t e r m s .$

$= [a + (2 - 1) d] + [a + (4 - 1) d] + - - - - - u p t o 100 t e r m s$

$= (a + d) + (a + 3 d) + - - - - - - u p t o 100 t e r m s$

$= (a + a + a + - - - - u p t o 100 t e r m s) + (d + 3 d + 5 d + - - - - u p t o 100 t e r m s)$

$= 100 a + d [\frac{100}{2} {2 (1) + (100 - 1) \times 2}]$

$S_{1} = 100 a + 10000 d . . . . . . . (1)$

So,

$S_{2} = a_{1} + a_{3} + - - - - - u p t o 100 t e r m s .$

$= [a + (1 - 1) d] + [a + (3 - 1) d] + - - - - - u p t o 100 t e r m s$

$= (a + 0 \times d) + (a + 2 d) + - - - - - - u p t o 100 t e r m s$

$= (a + a + a + - - - - u p t o 100 t e r m s) + (0 d + 2 d + 4 d + - - - - u p t o 100 t e r m s)$

$= 100 a + d [\frac{100}{2} {2 (0) + (100 - 1) \times 2}]$

$S_{1} = 100 a + 9900 d . . . . . . . (2)$

Then,

$S_{1} - S_{2} = 100 a + 10000 d - 100 a - 9900 d$

$= 100 d$

$S_{1} - S_{2} = 100 d$

$d = \frac{S_{1} - S_{2}}{100}$
Hence, this is the answer.

Suggest Corrections

Similar questions

The sums of n terms of three A.P.'s whose first term is 1 and common differences are 1, 2, 3 are $s_{1}, s_{2}, s_{3}$ respectively. The true relation is

The sum of n terms of three A.P.'s whose first term is 1 and

common differences are 1, 2, 3 are $S_{1}$ , $S_{2}$ , $S_{3}$ respectively. The true

relation is

If S denotes the sum of an infinite G.P. $S_{1}$ denotes the sum of the squares of its terms, then prove that the first term and common ratio are respectively $\frac{2 S S_{1}}{S^{2} + S_{1}}$ and $\frac{S^{2} - S_{1}}{S^{2} + S_{1}}$ .

If sum of n, 2n, 3n terms of an A.P. are $S_{1}, S_{2}, S_{3}$ respectively, then the value of $\frac{S_{3}}{S_{2} - S_{1}}$ is

Q. If

s_{1}, s_{2}, s_{3} . . . . . s_{q}

are the sums of n terms of q A.P.'s whose first terms are 1,2,3.....q and common differences are 1,3,5,....(2q - 1), respectively then

s_{1} + s_{2} + s_{3} . . . . s_{q} =

Join BYJU'S Learning Program

Ifs1=a2+a4+a6+.... upto 100 terms and s2=a1+a3+a5+.... upto 100 terms of a certain A.P., then its common difference is:

If $s_{1} = a_{2} + a_{4} + a_{6} + . . . .$ upto 100 terms and $s_{2} = a_{1} + a_{3} + a_{5} + . . . .$ upto 100 terms of a certain A.P., then its common difference is: