Q43 / 160 · AP EAMCET 2024 (20 May Shift 1) · Mathematics

$\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\ldots$ to 50 terms $=$

  • A
    $\frac{50}{203}$
  • Correct answer
    B
    $\frac{50}{609}$
  • C
    $\frac{150}{203}$
  • D
    $\frac{25}{609}$

Solution

$\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\ldots$ $T_n=\frac{1}{(4 n-1)(4 n+3)}=\frac{1}{4}\left[\frac{1}{4 n-1}-\frac{1}{4 n+3}\right]$ $T_1=\frac{1}{4}\left[\frac{1}{3}-\frac{1}{7}\right] \Rightarrow T_2=\frac{1}{4}\left[\frac{1}{7}-\frac{1}{11}\right]$ $T_{n-1}=\frac{1}{4}\left[\frac{1}{4 n-5}-\frac{1}{4 n-1}\right]$ $\mathrm{S}_{\mathrm{n}}=\mathrm{T}_1+\mathrm{T}_2+\ldots+\mathrm{T}_{\mathrm{n}}$ $=\frac{1}{4}\left[\frac{1}{3}-\frac{1}{4 n+3}\right]$ $S_n=\frac{n}{3(4 n+3)} \Rightarrow S_{50}=\frac{50}{3(200+3)}=\frac{50}{609}$

Hence, the correct answer is option: 'B'.

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