Question

if+the+first+&+the+nth+term+of+geometric+pprogression+are+a&b+&+if+P+is+the+product+of+are+first+n+terms,+prove+that+P+2+=(ab)n

Answer 1

The first term of the G.P is a and the last term is b.

Therefore, the G.P. is a, ar, ar2, ar3, … arn–1, where r is the common ratio.

b = arn–1 … (1)

P = Product of n terms

= (a) (ar) (ar2) … (arn–1)

= (a × a ×…a) (r × r2 × …rn–1)

= an r 1 + 2 +…(n–1) … (2)

Here, 1, 2, …(n – 1) is an A.P.

∴1 + 2 + ……….+ (n – 1)

equals fraction numerator n minus 1 over denominator 2 end fraction open square brackets 2 plus left parenthesis n minus 1 minus 1 right parenthesis cross times 1 close square brackets equals fraction numerator n minus 1 over denominator 2 end fraction open square brackets 2 plus n minus 2 close square brackets equals fraction numerator n open parentheses n minus 1 close parentheses over denominator 2 end fraction

P space equals space a to the power of n r to the power of fraction numerator n open parentheses n minus 1 close parentheses over denominator 2 end fraction end exponent

therefore space P squared space equals space a to the power of 2 n end exponent space r to the power of n open parentheses n minus 1 close parentheses end exponent

space space space space space space space space space space space space equals space open square brackets a squared space r to the power of left parenthesis n minus 1 right parenthesis end exponent close square brackets to the power of n

space space space space space space space space space space space space equals open square brackets a cross times a r to the power of n minus 1 end exponent close square brackets to the power of n

space space space space space space space space space space space space equals open parentheses a b close parentheses to the power of n space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets U sin g space left parenthesis 1 right parenthesis close square brackets

Thus, the given result is proved.