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JEE Main 20257 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2025 · 7 April · Evening Shift · Mathematics · Sequences and Series

Let a_ n be the n^ th term of an A. P.If S_n=a_1+a_2+a_3+ +a_n=700,...

Let an\mathrm{a}_{\mathrm{n}} be the nth \mathrm{n}^{\text{th }} term of an A. P.If Sn=a1+a2+a3+…+an=700,a6=7S_n=a_1+a_2+a_3+\ldots+a_n=700, a_6=7 and S7=7S_7=7, then an\mathrm{a}_{\mathrm{n}} is equal to :

Options

  1. A56
  2. B65
  3. C64
  4. D70

Correct answer

C. 64

Step-by-step solution

Sn=700=n2[2a+(n−1)d]\mathrm{S}_{\mathrm{n}}=700=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]...(i)
a6=7⇒a+5d=7a_6=7 \Rightarrow a+5 d=7...(ii)
S7=7⇒72(2a+6 d)=7\mathrm{S}_7=7 \Rightarrow \frac{7}{2}(2 \mathrm{a}+6 \mathrm{~d})=7
a+3d=1a+3 d=1...(iii)
Solve (ii) and (iii)
<br/>n2(−16+3n−3)=700⇒3n2−19n−1400=0(3n+56)(n−25)=0∴a25=a+24 d=−8+24×3=−8+72<br/>\begin{aligned}<br/>& \frac{\mathrm{n}}{2}(-16+3 n-3)=700 \Rightarrow 3 n^2-19 n-1400=0 \\ & (3 n+56)(\mathrm{n}-25)=0 \\ & \therefore \mathrm{a}_{25}=\mathrm{a}+24 \mathrm{~d}=-8+24 \times 3 \\ & =-8+72<br/>\end{aligned}
=64
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