If 10th term of an A.P is 0 and 42nd term is -32, then nth term is …

1.	If 10th term of an A.P is 0 and 42nd term is -32, then nth term is ……………… A) 10 – n B) 8 – n C) 11 – n D) 9 – n
Answer» Correct option is (A) 10 – n Let a & d be first term & common difference of A.P. We have \(a_{10}=0\;\&\;a_{42}=-32\) \(\therefore a+9d=0\) ______________(1) & \(a+41d=-32\) ______________(2) \((\because a_n=a+(n-1)d)\) Subtract equation (1) from (2), we get \((a+41d)-(a+9d)=-32-0\) \(\Rightarrow32d=-32\) \(\Rightarrow d=\frac{-32}{32}=-1\) \(\therefore a=-9d=-(-9)=9\) (From (1) & d = -1) \(\therefore a_n=a+(n-1)d\) \(=9+(n-1)d\) \(=9-n+1\) \(=10-n\) Hence, \(n^{th}\) term of A.P. is 10 - n. Correct option is A) 10 – n

If 10th term of an A.P is 0 and 42nd term is -32, then nth term is ……………… A) 10 – n B) 8 – n C) 11 – n D) 9 – n

Answer»

Correct option is (A) 10 – n

Let a & d be first term & common difference of A.P.

We have \(a_{10}=0\;\&\;a_{42}=-32\)

\(\therefore a+9d=0\) ______________(1)

& \(a+41d=-32\) ______________(2) \((\because a_n=a+(n-1)d)\)

Subtract equation (1) from (2), we get

\((a+41d)-(a+9d)=-32-0\)

\(\Rightarrow32d=-32\)

\(\Rightarrow d=\frac{-32}{32}=-1\)

\(\therefore a=-9d=-(-9)=9\) (From (1) & d = -1)

\(\therefore a_n=a+(n-1)d\)

\(=9+(n-1)d\)

\(=9-n+1\)

\(=10-n\)

Hence, \(n^{th}\) term of A.P. is 10 - n.

Correct option is A) 10 – n