Verified Solution • 2025–26 Session moderate QJEEM24Madb5e: JEE Main 2024 Mathematics - Class 11 JEEM24Madb5e Solution: Let denote the sum of first terms of an arithmetic progression. … Let SnS_nSn denote the sum of first nnn terms of an arithmetic progression. If S20=790S_{20}=790S20=790 and S10=145S_{10}=145S10=145 , then S15−S5\mathrm{S}_{15}-\mathrm{S}_5S15−S5 is : Multiple Choice Options A 405 B 390 C 410 D 395 Step-by-Step Verified Solution amp;amp;S20=202[2a+19 d]=790amp;2a+19 d=79.... (1)amp; S10=102[2a+9 d]=145amp;2a+9 d=29.... (2)amp; From (1) and (2) a=−8,d=5\begin{aligned}&\begin{aligned}& \mathrm{S}_{20}=\frac{20}{2}[2 \mathrm{a}+19 \mathrm{~d}]=790 \\& 2 \mathrm{a}+19 \mathrm{~d}=79 \quad \text{.... (1)}\\& \mathrm{~S}_{10}=\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}]=145 \\& 2 \mathrm{a}+9 \mathrm{~d}=29 \quad \text{.... (2)}\end{aligned}\\&\text { From (1) and (2) } a=-8, d=5\end{aligned}amp;amp;S20=220[2a+19 d]=790amp;2a+19 d=79.... (1)amp; S10=210[2a+9 d]=145amp;2a+9 d=29.... (2)amp; From (1) and (2) a=−8,d=5 amp;S15−S5=152[2a+14d]−52[2a+4d]amp;=152[−16+70]−52[−16+20]amp;=405−10amp;=395\begin{aligned}& S_{15}-S_5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d] \\& =\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20] \\& =405-10 \\& =395\end{aligned}amp;S15−S5=215[2a+14d]−25[2a+4d]amp;=215[−16+70]−25[−16+20]amp;=405−10amp;=395 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring