| The first term of an arithmetic progression is 22 and the last term is \[-11.\] If the sum is 66, the number of terms in the sequence are [SSC (10+2) 2014] |
A) 10
B) 12
C) 9
D) 8
Correct Answer: B
Solution :
| Given, \[a=22,\]\[l=-11,\]\[{{S}_{m}}=66\] |
| Let number of terms in sequence = n |
| Then, \[l=a+(n-1)\,d\] |
| \[\Rightarrow \] \[-11=22+(n-1)\,d\] |
| \[\Rightarrow \] \[(n-1)\,d=-\,22-11\] |
| \[\Rightarrow \] \[(n-1)\,d=-\,33\] ... (i) |
| Now, \[{{S}_{n}}=\frac{n}{2}\,\,[2a+(n-1)\,d]\] |
| \[\Rightarrow \] \[66=\frac{n}{2}\,\,[2\times 22+(n-1)\,d]\] |
| \[\Rightarrow \] \[66=\frac{n}{2}\,\,[44-33]\] |
| \[\Rightarrow \] \[66=\frac{n}{2}\times 11\]\[\Rightarrow \]\[n=\frac{66\times 2}{11}\] |
| \[\therefore \] \[n=12\] |
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