Question

The first and last term of a Geometrical Progression(G.P) are 3 and 96 respectively.If the common ratio is 2,find:
(i)'n' the number of terms of the G.P.

(ii) sum of the 'n' terms.​

Answer 1

(i) given first term(a)=3

last term(T)=96

common ratio(r)=2

last term in GP is ar^(n-1),n is total number of terms..hence,

96=(3)(2)^(n-1)

32=2^(n-1)

2^5=2^(n-1)

bases are equal..hence powers also equal

5=n-1

n=6 terms

(ii)sum of 'n' terms in GP is given by

S=a(r^n-1)/(r-1)

S=3(2^6-1)/(2-1)

S=3(64-1)/1

S=3(63)

S=189

3,6,12,24,48,96 are the numbers that are in GP

Answer 2

Answer:

(i)given first term(a)=3

last term(T)=96

common ratio(r)=2

last term in GP is ar^(n-1),n is total number of terms..hence,

96=(3)(2)^(n-1)

32=2^(n-1)

2^5=2^(n-1)

bases are equal..hence powers also equal

5=n-1

n=6 terms

(ii)sum of 'n' terms in GP is given by

S=a(r^n-1)/(r-1)

S=3(2^6-1)/(2-1)

S=3(64-1)/1

S=3(63)

S=189

3,6,12,24,48,96 are the numbers that are in GP

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