JEE Main202628 January 2026Evening ShiftMathematicsSequences and SeriesActual

6 3²⁶ + 10 1 3²⁵ + 10 2 3²⁴ + 10 2² 3²³ + + 10 2²⁴ 3 is equal to :

Options

A2²⁶
B3²⁵
C3²⁶
D2²⁵

Correct answer

A. 2²⁶

Step-by-step solution

The sum is 6 3²⁶ + _ r=0 ²⁴ 10 2^r 3^ 25-r = 6 3²⁶ + 10 1 3²⁵ _ r=0 ²⁴ (2 3)^r = 6 3²⁶ + 10 3²⁵ 6²⁵-1 5 = 6 + 2 3(6²⁵-1) 3²⁶ = 6 + 2 6²⁵ 3 - 6 3²⁶ = 6 6²⁵ 3²⁶ = 6²⁶ 3²⁶ = ( 6 3 )²⁶ = 2²⁶

Practice Sequences and Series on Quantrex Academy →

More from Sequences and Series

Let = 3+4+8+9+13+14+ upto 40 terms. If ( )^ 1020 is a root of the equation x^2+x-2=0 , (0, 2 ) , then ^2 + 3 ^2 is equal to: 2026 The sum 1 + 1 2 (1^2 + 2^2) + 1 3 (1^2 + 2^2 + 3^2) + upto 10 terms is equal to : 2026 The value of 1^3 - 2^3 + 3^3 - + 15^3 is: 2026 The sum of the first ten terms of an A.P. is 160 and the sum of the first two terms of a G.P. is 8 . If the first term of the A.P. is equal to the common ratio of the G.P. and the 2026 For the functions f( ) = ^2 + ^2 , and g( ) = ^2 + ^2 , > > 0 , let _ 0 < < /2 f( ) = _ 0 < < g( ) . If the first term of a G.P. is ( 2 ) , its common ratio is ( 2 ) and the sum of 2026 If the sum of the first 10 terms of the series 1 1 + 1^4 4 + 2 1 + 2^4 4 + 3 1 + 3^4 4 + 4 1 + 4^4 4 + is m n , (m, n) = 1 , then m + n is equal to : 2026 Let A₁, A₂, A₃, , A₃₉ be 39 arithmetic means between the numbers 59 and 159 . Then the mean of A₂₅, A₂₈, A₃₁ and A₃₆ is equal to : 2026 Let the sum of the first n terms of an A.P. be 3n^2 + 5n . Then the sum of squares of the first 10 terms of the A.P. is: 2026 Full Sequences and Series list All JEE Main PYQs