Question

Prove that the sum of any terms on an arithematic sequence is five times the middle term

Answer 1

SOLUTION :-

Let the 5 terms of the arithmetic sequence be $\sf a_1$ , $\sf a_2$ , $\sf a_3$ , $\sf a_4$ and $\sf a_5$ .

We know that,

 $\bf a_n=a+(n-1)d$

$\bullet \sf \ a_1=a+(1-1)d= a\\\\\bullet \ a_2=a+(2-1)d=a+d \\\\\bullet \ a_3=a+(3-1)d = a+2d \\\\\bullet \ a_4 = a+(4-1)d = a+3d \\\\\bullet \ a_5= a+(5-1)d=a+4d$

We have to prove that,

$\sf a_1+a_2+a_3+a_4+a_5= 5\times (a_3)$

L.H.S = $\sf a_1+a_2+a_3+a_4+a_5$

         = $\sf a + (a+d)+(a+2d)+(a+3d)+(a+4d)$

         = $\bf 5a+10d$

R.H.S = $\sf 5 \times (a_3)$

         = $\sf 5 \times (a+2d)$

         = $\bf 5a+10d$

$\bf \therefore L.H.S = R.H.S$

Hence proved.