Question

How many terms of the series $- 9, - 6, - 3,...$ must be taken that the sum may be 66?

Answer 1

Here, we can clearly see that the given series of numbers follow the Arithmetic progression. So, we have to use the concept of the Arithmetic progression (A.P.) as the series given is in AP form. Arithmetic progression (A.P.) is the sequence of numbers such that the difference between the consecutive numbers remains constant. So, we have to use the formula for the sum of n number of terms in an AP and equate it to 66. Then we will be able to find out the number of terms which should be added to get the sum of 66.

Formula used:
Sum of n number of terms in an AP $= \dfrac{{\rm{n}}}{2}\left[ {2{\rm{a}} + ({\rm{n}} - 1){\rm{d}}} \right]$

Complete step-by-step answer:
Series of number given is$- 9, - 6, - 3,...$
Here in this series of AP first term is $- 9$ and the common difference is 3 i.e. ${\rm{a}} = - 9,{\rm{d}} = 3$
Now it is given that the sum of the given series is 66. So, we have to equate the formula of the sum of n number of terms in an AP equals to 66 and then by solving we will get the value of n (number of terms).
Therefore, Sum of n number of terms in an AP $= \dfrac{{\rm{n}}}{2}\left[ {2{\rm{a}} + ({\rm{n}} - 1){\rm{d}}} \right]{\rm{ = 66}}$
Multiplying both side by 2 then the equation becomes
$\Rightarrow {\rm{n}}\left[ {2{\rm{a}} + ({\rm{n}} - 1){\rm{d}}} \right] = 66 \times 2$
Now, we have to put the value of a and d in the above equation, we get
$\Rightarrow {\rm{n}}\left[ {2( - 9) + ({\rm{n}} - 1)3} \right] = 132$
$\Rightarrow {\rm{n}}\left[ { - 18 + 3{\rm{n}} - 3} \right] = 132$
$\Rightarrow {\rm{n}}\left[ {3{\rm{n}} - 21} \right] = 132$
Now multiplying n with the bracket, we get
$\Rightarrow 3{{\rm{n}}^2} - 21{\rm{n}} = 132$
Now we have to solve this quadratic equation to get the value of n (number of terms).
$\Rightarrow 3{{\rm{n}}^2} - 33{\rm{n + 12n}} - 132 = 0$
$\Rightarrow 3{\rm{n(n}} - 11) + 12({\rm{n + 11)}} = 0$
By solving the quadratic equation we get the value of ${\rm{n = 11, - 4}}$
As the value of n cannot be negative so we will take the value of n as 11
Hence, 11 terms of the given series is added to get the sum 66.

Note: We always have to keep in mind the basic A.P. and G.P series
A.P. series is ${\rm{a}},{\rm{a}} + {\rm{d}},{\rm{a}} + 2{\rm{d}},{\rm{a}} + 3{\rm{d,}}............$
${{\rm{n}}^{th}}{\rm{term }} = {\rm{ a + }}\left( {{\rm{n - 1}}} \right){\rm{d}}$
Where a is the first term and d is the common difference
G.P. series is ${\rm{a}},{\rm{ar}},{\rm{a}}{{\rm{r}}^2},{\rm{a}}{{\rm{r}}^3},{\rm{a}}{{\rm{r}}^4},.........$
${{\rm{n}}^{{\rm{th}}}}{\rm{term = a}}{{\rm{r}}^{{\rm{n - 1}}}}$
Where, a is the first term of the G.P. and r is the common ratio.