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Find the sumof all multiples of 9 lying between 400 and 800

Posted by Tanu Rai 6 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 3 months ago

The numbers divisible by 9 between 400 and 800 are:

405, 414, 423,.................792

Here a=414, d=9 and l=792

Let the number of these terms be n, then

{tex}\mathrm{We}\;\mathrm{know}\;\mathrm{that}\;{\mathrm a}_{\mathrm n}=\mathrm a+(\mathrm n-1)\mathrm d{/tex}
a_n=792
Or, a + (n-1)d=792
{tex}\Rightarrow{/tex} 405 + (n-1){tex}\times{/tex}9=792
{tex}\Rightarrow{/tex} 9(n - 1) = 387
{tex}\Rightarrow{/tex} (n - 1) = 43
{tex}\Rightarrow{/tex} n = 44
So, S₄₄ = {tex}\frac{n}{2}{/tex}(a+l)
= {tex}\frac{{44}}{2}{/tex}(405 + 792)

= 22 {tex}\times{/tex} 1197 = 26334
Hence, S_n=26334

0Thank You

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CBSE > Class 10 > Mathematics