Question

Find the sum of first 12 terms of the series 2.5 + 5.8 + 8.11+...............is __

Answer 1

Observe that 2,5,8,11,--------are in A.P with first term 2 and common difference 3.

The first number of every term in this series froms A.P. with a=2, d=3. The second number of every

term in this series also forms A.P. with $a^{{}^{\prime }}$ = 2, $d^{{}^{\prime }}$ = 3

Hence $T_n$ = $[a+(n-1\left)d\right][a^{{}^{\prime }}+(n-1\left)d^{{}^{\prime }}\right]$

= $[2+(n-1\left)3\right][5+(n-1\left)3\right]$

= (3n - 1)(3n + 2)

Sum to 'n' terms:

Let $V_n$ = (3n - 1)(3n + 2)(3n + 5) -----------(1)

Then $V_{n-1}$ =(3n - 4)(3n - 1)(3n + 2)

$V_n$ - $V_{n-1}$ = (3n - 1)(3n + 2) [3n + 5 - 3n + 4]

= $T_n$.9

$\Rightarrow$ $T_n$ = $\frac{1}{9}$ ($V_n$ - $V_{n-1}$)

${\sum }_{n=1}^i{T}_n=\frac{1}{9}{\sum }_{n=1}^i(V_n-V_{n-1})$

${\sum }^i{T}_i=\frac{1}{9}[V_1-V_0+V_2-V_1+.........V_i-V_{i-1}]$

${\sum }^i{T}_i=\frac{1}{9}[V_i-V_0]$

We know from (1), $V_i$ = (3i - 1)(3i + 2)(3i + 5)

Hence , sum of 'i' terms is,

$S_i$ = $\frac{1}{9}$ [ (3i - 1)(3i + 2)(3i + 5) + 10 ]

So, sum of 12 terms is

$S_{19}$ = $\frac{1}{9}$ $[35\times 38\times 41+10]$

=6060