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JEE Main 202524 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2025 · 24 January · Evening Shift · Mathematics · Sequences and Series

In an arithmetic progression, if S_ 40 =1030 and S_ 12 =57 , then...

\text{ In an arithmetic progression, if } S_{40}=1030 \text{ and } S_{12}=57 \text{, then } S_{30}-S_{10} \text{ is equal to : }

Options

  1. A525
  2. B510
  3. C515
  4. D505

Correct answer

C. 515

Step-by-step solution

S40=1030⇒402[2a+39d]=1030⇒2a+39d=1032....(1)S12=57⇒122[2a+11d]=57⇒2a+11d=576...(2)\begin{aligned} & S_{40}=1030 \Rightarrow \frac{40}{2}[2 a+39 d]=1030 \\ & \Rightarrow \quad 2 a+39 d=\frac{103}{2}....(1) \\ & \quad S_{12}=57 \Rightarrow \frac{12}{2}[2 a+11 d]=57 \\ & \Rightarrow \quad 2 a+11 d=\frac{57}{6}...(2) \end{aligned} Equation (1) - equation (2) 28d=1032−57628d=309−576\begin{aligned} & 28 d=\frac{103}{2}-\frac{57}{6} \\ & 28 d=\frac{309-57}{6} \end{aligned}d=\frac{3}{2} \Rightarrow \quad a=-\frac{7}{2} S_{30}-S_{10}=\frac{30}{2}[2 a+29 d]-\frac{10}{2}[2 a+9 d] =15[2 a+29 d]-5[2 a+9 d] =5[6 a+87 d-2 a-9 d] =5[4 a+78 d] =5[-14+117] =515$
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