If the sum of the series 24, 20, 16, …………… is 60, then the

1.	If the sum of the series 24, 20, 16, …………… is 60, then the number of terms is A) 13 B) 25 C) 10D) 25
Answer» Correct option is (C) 10 24, 20, 16, …… is the arithmetic progression whose first term is \(a_1=a=24\;\&\) common difference is \(d=a_2-a_1\) \(=20-24=-4\) Let sum of n terms is 60. i.e., \(S_n=60\) \(\Rightarrow\frac n2[2a+(n-1)d]=60\) \(\Rightarrow n[2\times24+(n-1)\times-4]\) \(=60\times2=120\) \(\Rightarrow n(48-4n+4)=120\) \(\Rightarrow n(52-4n)=120\) \(\Rightarrow n(13-n)=\frac{120}4=30\) \(\Rightarrow n^2-13n+30=0\) \(\Rightarrow n^2-3n-10n+30=0\) \(\Rightarrow n(n-3)-10(n-3)=0\) \(\Rightarrow(n-3)(n-10)=0\) \(\Rightarrow n-3=0\;or\;n-10=0\) \(\Rightarrow n=3\;or\;n=10\) Hence, sum of first 3 terms or sum of first 10 terms of given sequence is 60. Correct option is C) 10

If the sum of the series 24, 20, 16, …………… is 60, then the number of terms is A) 13 B) 25 C) 10D) 25

Answer»

Correct option is (C) 10

24, 20, 16, …… is the arithmetic progression whose first term is \(a_1=a=24\;\&\)

common difference is \(d=a_2-a_1\)

\(=20-24=-4\)

Let sum of n terms is 60.

i.e., \(S_n=60\)

\(\Rightarrow\frac n2[2a+(n-1)d]=60\)

\(\Rightarrow n[2\times24+(n-1)\times-4]\)

\(=60\times2=120\)

\(\Rightarrow n(48-4n+4)=120\)

\(\Rightarrow n(52-4n)=120\)

\(\Rightarrow n(13-n)=\frac{120}4=30\)

\(\Rightarrow n^2-13n+30=0\)

\(\Rightarrow n^2-3n-10n+30=0\)

\(\Rightarrow n(n-3)-10(n-3)=0\)

\(\Rightarrow(n-3)(n-10)=0\)

\(\Rightarrow n-3=0\;or\;n-10=0\)

\(\Rightarrow n=3\;or\;n=10\)

Hence, sum of first 3 terms or sum of first 10 terms of given sequence is 60.

Correct option is C) 10