Question

Find the sum of the series 3.75,3.5,3.25……….upto 16 terms.

Answer 1

Hint- First find out the type of progression which the sequence is in that is if it is in A.P, G.P or H.P and solve it.

The series given to us is 3.75,3.5,3.25…………..
We have been asked to find out the sum of the series upto 16 terms
From the series given we get $T_2-T_1=3.5-3.75=-0.25$
                         Also, we get $T_3-T_2=3.25-3.50=-0.25$
So, from this we got $T_3-T_2=T_2-T_1$ =common difference=d
So, from this we can conclude that the given series is in Arithmetic Progression(A.P)
So, we know that the sum of n terms of an A.P is given by
$S_n={\displaystyle \frac{n}{2}}\left[2a+\left(n-1\right)d\right]$
So, on comparing with the sequence ,we can write
The first term=a=3.75
Common difference d=-0.25
Here, since we have to find out the sum upto 16 terms, we consider n=16
Let us substitute these values in the $S_n$ formula
So, we get $$\begin{gathered} S_{16}={\displaystyle \frac{16}{2}}\left(2\times 3.75+(16-1)(-0.25)\right) \\ {S}_{16}=8(7.5-3.75) \\ {S}_{16}=8(3.75) \\ \Rightarrow S_{16}=30 \end{gathered}$$
So, the sum of the series upto 16 terms=30

Note: When finding sum to n terms of an AP we can make use of an alternative formula if the first and last terms of an AP are known or we can use the same formula as used in this problem and solve.