Verified Solution • 2025–26 Session moderate

QJEEM25Mad54d: JEE Main 2025 Mathematics - Class 11 JEEM25Mad54d Solution: Let be the term of an A.P. If for some , and , then is equal to

Let Tr\mathrm{T}_{\mathrm{r}} be the rth \mathrm{r}^{\text {th }} term of an A.P. If for some m,Tm=125, T25=120\mathrm{m}, \mathrm{T}_{\mathrm{m}}=\frac{1}{25}, \mathrm{~T}_{25}=\frac{1}{20} , and 20∑r=125 Tr=1320 \sum\limits_{\mathrm{r}=1}^{25} \mathrm{~T}_{\mathrm{r}}=13 , then 5 m∑r=m2 m Tr5 \mathrm{~m} \sum\limits_{\mathrm{r}=\mathrm{m}}^{2 \mathrm{~m}} \mathrm{~T}_{\mathrm{r}} is equal to

Multiple Choice Options

  • A 98
  • B 126
  • C 112
  • D 142

Step-by-Step Verified Solution

amp;Tm=125, T25=120,20∑r=125 Tr=13amp; Tm=a+(m−1)d=125…….(1)amp;T25=a+24 d=120\begin{aligned}& \mathrm{T}_{\mathrm{m}}=\frac{1}{25}, \mathrm{~T}_{25}=\frac{1}{20}, 20 \sum_{\mathrm{r}=1}^{25} \mathrm{~T}_{\mathrm{r}}=13 \\& \mathrm{~T}_{\mathrm{m}}=\mathrm{a}+(\mathrm{m}-1) \mathrm{d}=\frac{1}{25} \ldots \ldots .(1) \\& \mathrm{T}_{25}=\mathrm{a}+24 \mathrm{~d}=\frac{1}{20}\end{aligned}

20.252[a+120]=13⇒a=150020.\frac{25}{2}\left[a+\frac{1}{20}\right]=13 \Rightarrow a=\frac{1}{500}

also, 20 S25=20⋅252[2a+24 d]=13⇒ d=150020 \mathrm{~S}_{25}=20 \cdot \frac{25}{2}[2 \mathrm{a}+24 \mathrm{~d}]=13 \Rightarrow \mathrm{~d}=\frac{1}{500}

from (1) 1500+m−1500=125⇒m=20\frac{1}{500}+\frac{m-1}{500}=\frac{1}{25} \Rightarrow m=20

Now,

5 m∑r=m2 m Tr=100∑r=2040 Tr=126

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