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Sequences and Series · Question 21

JEE Main 2026 (Online) 28th January Evening ShiftNumericalMedium📈 Topic in 23% papers+4
1

If r=125 ( rr4 + r2 + 1 ) = pq, where p and q are positive integers such that (p, q) = 1, then p + q is equal to .

Answer

976

Solution

r4 + r2 + 1 = (r2 + 1)2 - r2 = (r2 - r + 1)(r2 + r + 1)

rr4 + r2 + 1 = 12 [ 1r2 - r + 1 - 1r2 + r + 1 ]

(r + 1)2 - (r + 1) + 1 = r2 + r + 1, the sum telescopes:

r=125 = 12 [1 - 1252 + 25 + 1 ] = 12 [1 - 1651 ]

= 6501302 = 325651

325 = 52 13, 651 = 3 7 31 (325, 651) = 1

p + q = 325 + 651 = 976

Answer: 976

JEE Main 2026 (Online) 28th January Evening Shift

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