Q. 133.6( 5 Votes )

# The sum of an infinite GP is 57, and the sum of their cubes is 9747. Find the GP.

Answer :

Let the first term Of G.P. be a, and common ratio be r.

∴

On cubing each term will become,

a^{3,} a^{3}r^{3}, ….

∴This sum

a=57(1-r) put this in equation 2 we get

⇒

⇒

⇒ 19(1-2r+r^{2})=1+r+r^{2}

⇒ 19r^{2}-r^{2}-38r-r+19-1=0

⇒ 18r^{2}-39r+18=0

⇒ 6r^{2}-13r+6=0

⇒ (2r-3)(3r-2)=0

⇒ r= 2/3, 3/2

But -1<r<1

⇒ r=2/3

Substitute this value of r in equation 1 we get

Thus the first term of G.P. is 19, and the common ratio is 2/3

∴G.P=19,

19,

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The sum of n terms of a progression is (2^{n} – 1). Show that it is a GP and find its common ratio.

If a, b, c, d are in GP, prove that

(i) (b + c)(b + d) = (c + a)(c + a)

(ii)

(iii) (a + b + c + d)^{2} = (a + b)^{2} + 2(b + c)^{2} + (c + d)^{2}

RS Aggarwal - Mathematics

Evaluate :

NOTE:In an expression like this ⇒ , n represents the upper limit, 1 represents the lower limit , x is the variable expression which we are finding out the sum of and i represents the index of summarization.

(i)

(ii)

(iii)

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The first term of a GP is 27, and its 8^{th} term is . Find the sum of its first 10 terms.