Question

(0.3)1+(0.3)2+(0.3)3+(0.3)4................+(0.3)40=
Please share the formalae that you used to solve it

Answer 1

Answer:

0.3(1+2+3.....40)

0.3( 20(2+ 39))

0.3(20*41)

0.3(820)

246

s= n/2(2a + (n-1)d)

Answer 2

$\:\:\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge\mathcal{\bf{\underline{\underline{\huge\mathcal{Answer}}}}}$

$\underline{\large\mathcal\red{solution}}$

$\underline{\large\mathcal\red{....formula\: used....}}$

if ,a and l be the first and the last term of an AP series...then ..sum is...

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

$\underline{\large\mathcal\red{....given\: AP \: series\: is....}}$

(0.3)1+(0.3)2+(0.3)3+(0.3)4................+(0.3)40

here,,,,[a=0.3]......[l=(0.3)40].....n=40

therefore ...the sum is..,

$s(40) = \frac{40(0.3 + (0.3)40)}{2} \\ = > s(40) = \frac{0.3 \times 40 \times 41}{2} \\ = >s(40) = 240$

$\large\mathcal\red{hope\: this \: helps \:you......}$