Question

Find the sum of the AP 4,8,12,16, upto 40

Answer 1

Given

A.P. — 4 , 8 ,12 ,16,...40

Here ,

• a ( first term ) → 4
• d ( common difference ) → 4
• l ( last term ) → 40

Let, number of terms be n .

Now,

$\bold{★ \: \: \: \: l \: = a \: + (n - 1)d}$

$\implies \: 40 = 4 + (n - 1)4$

$\implies \: 40 - 4 =(n - 1)4$

$\implies \: 36 =(n - 1)4$

$\implies \: n - 1 = \frac{36}{4}$

$\implies \: n - 1 = 9$

$\implies \: n = 9 + 1$

$\implies \: \underline \bold{n = 10}$

Sum of the A.P. →

$\bold{S_n = \frac{n}{2}(a + l)}$

$\implies \bold{ S_{10 }= \frac{10}{2} (4 + 40)}$

$\implies \bold{ S_{10 }= \frac{10}{2} \times 44}$

$\implies \bold{ S_{10 }= {10} \times 22}$

$\implies \boxed{\bold{ S_{10 }= 220}}$

$\Large{\colorbox{yellow}{\underline{\underline{♠Answer♠:—\:\:220} }}}$