JEE MathsSequence & SeriesVisual Solution
Visual SolutionPYQ 2023 · Apr 10 Shift 1Standard
Question
Let the first term aa and the common ratio rr of a geometric progression be positive integers. If the sum of squares of its first three terms is 3303333033, then the sum of these three terms is equal to
(A)241241
(B)231231
(C)210210
(D)220220
Solution Path
a2(1+r2+r4)=33033a^2(1 + r^2 + r^4) = 33033. With r=4r = 4, a=11a = 11. Sum =11(21)=231= 11(21) = 231.
01Question Setup
1/4
GP with positive integer aa and rr. Sum of squares of first 3 terms is 3303333033. Find the sum.
a+ar+ar2=  ?a + ar + ar^2 = \;?
02Set Up Equation
2/4
a2(1+r2+r4)=33033a^2(1 + r^2 + r^4) = 33033. Since a,ra, r are positive integers, try values of rr.
a2(1+r2+r4)=33033a^2(1 + r^2 + r^4) = 33033
03Find a and rKEY INSIGHT
3/4
Try r=4r = 4: 1+16+256=2731 + 16 + 256 = 273. Then a2=33033/273=121a^2 = 33033/273 = 121, so a=11a = 11. Sum =11(1+4+16)=231= 11(1 + 4 + 16) = 231.
11×21=23111 \times 21 = 231
04Final Answer
4/4
Sum of three terms =231= 231.
231\boxed{231}