Home › UP Board Class 10 › Mathematics › How many terms of the AP $9, 17, 25, \ldots$ mus…
How many terms of the AP $9, 17, 25, \ldots$ must be taken to give a sum of $636$?
A10
B11
C12
D13
Answer & Solution
Correct answer: C. 12
Here $a=9$ and $d=8$. Using $S_n=\frac{n}{2}[2a+(n-1)d]$, we get
$$636=\frac{n}{2}[18+8(n-1)]=\frac{n}{2}(8n+10)=n(4n+5).$$
So $4n^2+5n-636=0$. Solving gives $n=12$ (the other root is negative, so it is rejected).
Related questions
When variables and numbers appear on both sides, you must simplify:Equations needing more than one operation are described as taking more:Once a solution is found, it should be checked by substituting it back to get a statement The product of any number and its reciprocal equals:A fraction multiplying the variable is best removed by multiplying by its:Solving x divided by 4 equals 3 gives x equal to:Solving 4x equals 20 gives x equal to:Solving x minus 8 equals 5 gives x equal to: