For the following G-P-s- find Sni 2-6-18-54-ii a-b-b2-a- b3-a2-

S_n=a(r^n 1)r 1 a) S_n=2(3^n 1)(3 1)=3^n 1 b) S_n=a[(b/a)^n 1](ba 1)=a^2[(b/a)^n 1](b a) . ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Sum of Coefficients of All Terms

for the follo...

Question

for the following G.P.s. find $S_{n}$
i) $2, 6, 18, 54, . . . . . .$
ii) $a, b, \frac{b^{2}}{a}, \frac{b^{3}}{a^{2}} . . . .$

Open in App

Solution

$S_{n} = \frac{a (r^{n} - 1)}{r - 1}$
a) $S_{n} = \frac{2 (3^{n} - 1)}{(3 - 1)} = 3^{n} - 1$
b) $S_{n} = \frac{a [(b / a)^{n} - 1]}{(\frac{b}{a} - 1)} = \frac{a^{2} [(b / a)^{n} - 1]}{(b - a)}$ .

Suggest Corrections

Similar questions

Q. For the following

G . P . s,

find

S_{n}

2, 6, 18, 54, . . .

Q. Simplify the following expressions.

\frac{a^{2} - b^{2}}{a - b} - \frac{a^{3} - b^{3}}{a^{2} - b^{2}}

Expand the following formulas

(a + b)^{2}

(a - b)^{2}

(a^{2} + b^{2})

(a + b)^{3}

(a - b)^{3}

a^{3} + b^{3}

a^{3} - b^{3}

Q. Which of the following equations are correct?

(i) (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)

(i i) (a + b)^{3} = a^{3} + b^{3} + 3 a b (a - b)

(i i i) (a - b)^{3} = a^{3} - b^{3} - 3 a b (a + b)

(i v) a^{2} - b^{2} = (a + b) (a - b)

Which of the following equation/s is/are correct?

$i) (a + b + c)^{2} = a^{2} + b^{2} + c^{2} - 2 (a b + b c + c a)$
$i i) (a + b)^{3} = a^{3} + b^{3} + 3 a b (a - b)$
$i i i) (a - b)^{3} = a^{3} - b^{3} - 3 a b (a + b)$
$i v) a^{2} - b^{2} = (a + b) (a - b)$

Join BYJU'S Learning Program

for the following G.P.s. find Sni) 2,6,18,54,......ii) a,b,b2a,b3a2....

Sn=a(rn−1)r−1a) Sn=2(3n−1)(3−1)=3n−1b) Sn=a[(b/a)n−1](ba−1)=a2[(b/a)n−1](b−a).

for the following G.P.s. find $S_{n}$
i) $2, 6, 18, 54, . . . . . .$
ii) $a, b, \frac{b^{2}}{a}, \frac{b^{3}}{a^{2}} . . . .$

$S_{n} = \frac{a (r^{n} - 1)}{r - 1}$
a) $S_{n} = \frac{2 (3^{n} - 1)}{(3 - 1)} = 3^{n} - 1$
b) $S_{n} = \frac{a [(b / a)^{n} - 1]}{(\frac{b}{a} - 1)} = \frac{a^{2} [(b / a)^{n} - 1]}{(b - a)}$ .