Question

n arithmetic means are inserted between 7 and 49. If sum of these means is 364 then sum of there square is

Answer 1

Answer:

11830

Step-by-step explanation:

Given: n arithmetic means are inserted between 7 and 49.

The sequence can be written as:

7, a+d, a+2d,...........a+nd, 49, where a+d, a+2d,...........a+nd=364 and a=7.

Number of terms=n+2, a=7 and l=49 and $s_{n} =364$

Now, if  $s_{n} =364$,  then $s_{n+2} =364+7+49$=$420$

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

⇒$s_{n+2}=\frac{n+2}{2} (a+l)$

⇒$420=\frac{n+2}{2}(7+49)$

⇒$840=7n+14+49n+98$

⇒$n=13$

Now, 49=7+(n+2-1)d

⇒d=3 (By putting n=13)

Also, S=$(a+d)^{2}+(a+2d)^{2}  ............(a+nd)^{2}$

⇒S=$na^{2} +d^{2} [1+4+9+......n^{2}]+2ad[1+2+3+....+n]$

⇒S=$na^{2} +d^{2} \frac{(n)(n+1)(2n+1)}{6}+2ad\frac{n(n+1)}{2}$

⇒S=$13(49)+9[\frac{(13)(14)(27)}{6}] +[\frac{(2)(7)(3)(13)(14)}{2}]$

⇒S=$637+7371+3822$

⇒S=$11830$

Answer 2

Answer:

13 = n , the sum of the squares = 11830

Step-by-step explanation:

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