JEE Main 26 Feb 2021 Shift 2 Solved Paper

If AM and GM satisfy the equation 4x2−9x+5=0, then AM and GM are nothing but roots of this quadratic equation,
4x2−9x+5=0
⇒‌‌4x2−4x−5x+5‌‌=0
⇒‌‌4x(x−1)−5(x−1)‌‌=0
⇒‌‌(x−1)(4x−5)‌‌=0⇒x=1,‌

5

4

M=1 [∵AM≥GM]
Then, AM=‌

5

4

and GM=
Again, the given series is
−16,8,−4,2......
which is a geometric progression series with common ratio ‌

−1

2

, then p th term =−16(‌

−1

2

)p−1=tp
q th term =−16(‌

−1

2

)q−1=tq
Arithmetic mean =‌

5

4

⇒‌‌‌

tp+tq

2

=‌

5

4

and Geometric mean =1
‌⇒‌√tptq‌=1
‌∵‌tptq‌=1
⇒‌(−16)(‌

−1

2

)p−1(−16)(‌

−1

2

)q−1‌=1
⇒‌(−16)2(‌

−1

2

)p+q−2‌=1
⇒‌(−24)2(‌

−1

2

)p+q−2‌=1
⇒‌(−2)8‌

(1)p+q−2

(−2)p+q−2

‌=1
⇒‌(−2)8‌=(−2)p+q−2
‌(−2)8‌=(−2)p+q−2
⇒‌p+q−2‌=8
⇒‌p+q‌=10