Evaluate `9 + 99 + 999 +………`upto n terms.

Evaluate: (999)^3

0.9 + 0.99 + 0.999 + ……. up to 51 terms = 51 - (1)/(p)(1-(1))/(10^(q)) where p, q in N then find the value (p + q)/(15) .

Evaluate: -999 div (-1)

Evalute: (Without expanding at any stage) |(a,2b,4c),(2,2,2),(2^(n),3^(n),5^(n))| , where a = 1 + 2 + 4 + …. upto n terms, b= 1 + 3 + 9 + …. upton n terms and c=1 +5+25 …upto n terms.

If S_(n) = 1 + 3 + 7 + 13 + 21 + "….." upto n terms, then

Find the sum of n terms of series : 3 + 5 + 9 + 15 + 23 + ………… n terms

Find the sum of G.P. : (x+y)/(x-y)+1+(x-y)/(x+y)+ . . . . .. . . . upto n terms.

If (1 + 3 + 5 + .... " upto n terms ")/(4 + 7 + 10 + ... " upto n terms") = (20)/(7 " log"_(10)x) and n = log_(10)x + log_(10) x^((1)/(2)) + log_(10) x^((1)/(4)) + log_(10) x^((1)/(8)) + ... + oo , then x is equal to

Find the sum of the geometric series : 1,(1)/(2),(1)/(4),(1)/(8), . . . .. . . . . . upto 12 terms.

Each of two arithmetic progressions 2, 4, 6, ... and 3, 6, 9, ... are taken upto 200 terms. How many terms are common in these two progressions ?