Question

the sum of the first term and fifth term term of an ascending AP is 26 and the product of the second term by fourth term is 160 find the sum of the first term seven terms of this AP

Answer 1

for this d=4√6 and a= 8 [2-√6]

Answer 2

Answer:

The sum of the first seven terms of the given AP is 112.

Solution:

Let a = the first term of an AP, n = nth term, d = common difference.

So, 1st term of an Arithmetic Progression $t_1$ = a    

And, 5th term of an Arithmetic Progression $t_5$ = a + (5 -1) d = a + 4d  

Given,  

$\begin{array} { l } { \mathrm { t } _ { 1 } + \mathrm { t } _ { 5 } = 26 } \\\\ { \mathrm { a } + \mathrm { a } + 4 \mathrm { d } = 26 } \\\\ { 2 \mathrm { a } + 4 \mathrm { d } = 26 } \\\\ { 2 ( \mathrm { a } + 2 \mathrm { d } ) = 26 } \\\\ { \mathrm { a } + 2 \mathrm { d } = \frac { 26 } { 2 } } \\\\ { \mathrm { a } + 2 \mathrm { d } = 13 } \\\\ { \mathrm { a } = 13 - 2 \mathrm { d } \ldots . . \text { (i) } } \end{array}$

2nd term of an Arithmetic Progression $t_2$ = a + d  

4th term of an Arithmetic Progression $t_4$ = a + (4-1) d = a + 3d  

Given,  

$\begin{array} { l } { \mathrm { t } _ { 2 } \times \mathrm { t } _ { 4 } = 160 } \\\\ { ( \mathrm { a } + \mathrm { d } ) ( \mathrm { a } + 3 \mathrm { d } ) = 160 } \\\\ { a ^ { 2 } + a d + 3 a d + 3 d ^ { 2 } = 160 } \end{array}$

Putting a = 13 – 2d from (i) in the above equation, we get,

$\begin{array} { l } { ( 13 - 2 d ) ^ { 2 } + ( 13 - 2 d ) d + 3 ( 13 - 2 d ) d + 3 d ^ { 2 } = 160 } \\\\ { 169 - 52 d + 4 d ^ { 2 } + 13 d - 2 d ^ { 2 } + 39 d - 6 d ^ { 2 } + 3 d ^ { 2 } = 160 } \\\\ { 169 - 52 d + 13 d + 39 d + 4 d ^ { 2 } + 3 d ^ { 2 } - 2 d ^ { 2 } - 6 d ^ { 2 } = 160 } \\\\ { 169 - d ^ { 2 } = 160 } \\\\ { 169 - 160 - d ^ { 2 } = 0 } \\\\ { 9 - d ^ { 2 } = 0 } \\\\ { d ^ { 2 } = 9 } \\\\ { d = \sqrt { 9 } } \end{array}$

d= +3,-3  

Since, it Is an ascending AP, d = +3.

So,  

a = 13 – (2 x 3) = 13 – 6 = 7

Hence, the sum of the first seven terms of this ascending AP is

$\begin{aligned} = \frac { 7 } { 2 } & ( 2\times7 + ( 7 - 1 ) 3 ) \\\\ & = \frac { 7 } { 2 } ( 14 + 6\times3 ) \\\\ & = \frac { 7 } { 2 } ( 14 + 18 ) \\\\ & = \frac { 7 } { 2 } \times 32 \\\\ & = 112 \end{aligned}$

The sum of the first seven terms of the given AP is 112.