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QJEEM24Ma7198: JEE Main 2024 Mathematics - Class 11 JEEM24Ma7198 Solution: Let the first three terms 2, p and q, with , of a G.P. be…

Let the first three terms 2, p and q, with q≠2q \neq 2 , of a G.P. be respectively the 7th ,8th 7^{\text {th }}, 8^{\text {th }} and 13th 13^{\text {th }} terms of an A.P. If the 5th 5^{\text {th }} term of the G.P. is the nth n^{\text {th }} term of the A.P., then nn is equal to:

Multiple Choice Options

  • A 151
  • B 177
  • C 163
  • D 169

Step-by-Step Verified Solution

amp; Let p=2r,q=2r2amp;T7=2,T8=2r,T13=2r2amp;d=2r−2=2(r−1)amp;2r2=T7+6d=2+6(2)(r−1)=12r−10amp;⇒r2−6r+5=0amp;⇒(r−1)(r−5)=0amp;∴r=1,5amp;r=1 (rejected) as q≠2amp;∴r=5\begin{aligned}& \text { Let } p=2 r, q=2 r^2 \\& T_7=2, T_8=2 r, T_{13}=2 r^2 \\& d=2 r-2=2(r-1) \\& 2 r^2=T_7+6 d=2+6(2)(r-1)=12 r-10 \\& \Rightarrow r^2-6 r+5=0 \\& \Rightarrow(r-1)(r-5)=0 \\& \therefore r=1,5 \\& r=1 \text { (rejected) as } q \neq 2 \\& \therefore r=5\end{aligned}

5th 5^{\text {th }} term of G.P =2.r4=2.54=2 . r^4=2.5^4

Let 1st 1^{\text {st }} term of A.P ba=a,d=8b a=a, d=8

2=a+(6)(8)⇒a=−462=a+(6)(8) \Rightarrow a=-46

nth \mathrm{n}^{\text {th }} term of A.P =−46+(n−1)8=8n−54=-46+(n-1) 8=8 n-54

amp;2.54=8n−54amp;⇒1250+54=8namp;⇒n=13048=163\begin{aligned}& 2.5^4=8 n-54 \\& \Rightarrow 1250+54=8 n \\& \Rightarrow n=\frac{1304}{8}=163\end{aligned}

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