Question

find the sum of the first 40 terms of the ap whose 4th term is 8 and 6th term is 14​

Answer 1

Given:-

• 4th term of an A.P = 8

• 6th term of an A.P = 14

• No. of term = 40

To Find:-

• Sum of first 49 terms

Formula Used:-

• ${\boxed{\bf{a_n=a+(n-1)d}}}$

• ${\boxed{\bf{S_n=\dfrac{n}{2}[2a+(n-1)d]}}}$

Solution:-

As given,

$\sf :\implies\:a_4= 8$

$\sf :\implies\:a+(n-1)d= 8$

$\sf :\implies\:a+(4-1)d= 8$

$\sf :\implies\:a+3d= 8$

$\sf :\implies\:a= 8-3d$ ____{1}

And,

$\sf :\implies\:a_6= 14$

$\sf :\implies\:a+(n-1)d= 14$

$\sf :\implies\:a+(6-1)d= 14$

$\sf :\implies\:a+5d= 14$

Putting Value of a,

$\sf :\implies\:8-3d+5d= 14$

$\sf :\implies\:8+2d= 14$

$\sf :\implies\:2d= 22$

$\sf :\implies\:d=\dfrac{22}{2}$

$\bf :\implies\:d= 11$

Putting value of d in {1},

$\sf :\implies\:a= 8-3\times11$

$\sf :\implies\:a= 8-33$

$\bf :\implies\:a= -25$

Now, Sum of first 40 terms:-

$\sf :\implies\:S_n=\dfrac{n}{2}[2a+(n-1)d]$

$\sf :\implies\:S_{40}=\dfrac{40}{2}[2\times(-25)+(40-1)11]$

$\sf :\implies\:S_{40}=20[-50+39\times11]$

$\sf :\implies\:S_{40}=20[-50+429]$

$\sf :\implies\:S_{40}=20\times379$

$\bf :\implies\:S_{40}=7580$

Hence, The sum of first 40 terms of an A.P is 7,580.

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