the sum of the first seven …

Let a be the first term and d be the common difference of the given AP.

Now a₄=a+(4-1)d

{tex}\implies{/tex}a₄=a+3d.

And, a₁₇=a+(17-1)d

{tex}\implies{/tex}a₁₇= a+16d.
Then,
T₄=a+3d and T₁₇= a+16d
Now, {tex}\frac { T _ { 4 } } { T _ { 17 } } = \frac { 1 } { 5 }{/tex}
{tex}\Rightarrow \frac { a + 3 d } { a + 16 d } = \frac { 1 } { 5 }{/tex}
{tex}\Rightarrow{/tex} 5a+15d=a+16d

{tex}\implies{/tex}5a-a+15d-16d=0
{tex}\Rightarrow{/tex} 4a-d=0
{tex}\Rightarrow{/tex} 4a=d....(i)
Also, S₇=182

Where ,S₇={tex}\frac {7}2 [{2a+(7-1)d}]{/tex}
{tex}\Rightarrow \frac { 7 } { 2 } [ 2 a + 6 d ] = 182{/tex}
{tex}\Rightarrow \frac { 7 \times 2 } { 2 } [ a + 3 d ] = 182{/tex}
{tex}\Rightarrow{/tex} a+3d=26
{tex}\Rightarrow{/tex} a+3(4a)=26...[from (i)]
{tex}\Rightarrow{/tex} 13a=26
{tex}\Rightarrow{/tex} a=2
{tex}\Rightarrow{/tex} d=4(2)=8
Thus, We have
T₁ = 2
T₂ = T₁ + d = 2 + 8 = 10
T₃ = T₁ + 2d = 2 + 2(8) = 2 + 16 = 18
T₄ = T₁ + 3d = 2 + 3(8) = 2 + 24 = 26
Thus, the given AP is 2,10,18,26,...