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BITSAT 2021
Exam
BITSAT
Year
2021
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2021 · Mathematics · Sequences and Series

If the sum of a certain number of terms of the A.P. 25,22,19, is...

If the sum of a certain number of terms of the A.P. 25,22,19,…25,22,19, \ldots is 116 then the last term is

Options

  1. A0
  2. B2
  3. C4
  4. D6

Correct answer

C. 4

Step-by-step solution

a=25,d=22−25=−3. Leta=25, d=22-25=-3 \text{. } Letnbetheno.oftermsbe the no. of terms Sum =116; Sum =n2[2a+(n−1)d]116=n2[50+(n−1)(−3)] or 232=n[50−3n+3]=n[53−3n]=−3n2+53n⇒3n2−53+232=0⇒(n−8)(3n−29)=0⇒n=8 or n=293,n≠293∴n=8∴ Now, T8=a+(8−1)d=25+7×(−3)=25−21=4∴ Last term =4\begin{array}{l} \text{ Sum }=116 ; \\ \text{ Sum }=\frac{n}{2}[2 a+(n-1) d] \\ 116=\frac{n}{2}[50+(n-1)(-3)] \\ \text{ or } 232=n[50-3 n+3]=n[53-3 n]=-3 n^{2}+53 n \\ \Rightarrow 3 n^{2}-53+232=0 \\ \Rightarrow(n-8)(3 n-29)=0 \\ \Rightarrow n=8 \text{ or } n=\frac{29}{3}, \\ n \neq \frac{29}{3} \\ \therefore n=8 \\ \therefore \text{ Now, } \mathrm{T}_{8}=a+(8-1) d=25+7 \times(-3)=25-21=4 \\ \therefore \text{ Last term }=4 \end{array}$$
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