Question

the second and the fourth teram of an A.P are 12 and 20 respectively find the sum of the frist 25 terms of that A.P​

Answer 1

S O L U T I O N :

$\underline{\bf{Given\::}}$

The second & the fourth terms of an A.P. are 12 & 20 respectively.

$\underline{\bf{Explanation\::}}$

As we know that formula of an A.P;

$\boxed{\bf{a_n =a+(n-1)d}}$

• a is the first term.
• d is the common difference.
• n is the term of an A.P.

$\underline{\underline{\tt{According\:to\:the\:question\::}}}$

$\mapsto\tt{a_2 = 12}$

$\mapsto\tt{a+(2-1)d = 12}$

$\mapsto\tt{a+(1)d = 12}$

$\mapsto\tt{a+d= 12}$

$\mapsto\tt{a= 12-d............(1)}$

&

$\mapsto\tt{a_4 = 20}$

$\mapsto\tt{a+(4-1)d = 20}$

$\mapsto\tt{a+(3)d= 20}$

$\mapsto\tt{a+3d= 20}$

$\mapsto\tt{(12-d)+3d= 20\:\:[from(1)]}$

$\mapsto\tt{12-d+3d= 20}$

$\mapsto\tt{12+2d= 20}$

$\mapsto\tt{2d= 20-12}$

$\mapsto\tt{2d= 8}$

$\mapsto\tt{d= \cancel{8/2}}$

$\mapsto\bf{d = 4}$

∴ Putting the value of d in equation (1),we get;

$\mapsto\tt{a = 12-4}$

$\mapsto\bf{a = 8}$

Now,

As we know that formula of the sum of an A.P.;

$\boxed{\bf{S_n = \frac{n}{2} \bigg[2a+(n-1)d\bigg]}}$

$\longrightarrow\tt{S_{25} = \dfrac{25}{2} \bigg[2(8) + (25-1)(4)\bigg]}$

$\longrightarrow\tt{S_{25} = \dfrac{25}{2} \bigg[16 + (24)(4)\bigg]}$

$\longrightarrow\tt{S_{25} = \dfrac{25}{2} \bigg[16 + 96\bigg]}$

$\longrightarrow\tt{S_{25} = \dfrac{25}{2} \bigg[112\bigg]}$

$\longrightarrow\tt{S_{25} = \dfrac{25}{\cancel{2}} \times \cancel{112}}$

$\longrightarrow\tt{S_{25} = 25 \times 56}$

$\longrightarrow\bf{S_{25} = 1400}$

Thus,

The sum of the first 25 terms of that A.P. will be 1400 .