Question

The sum of the first 5 and first 10 terms respectively of an A.P. Are equal in magnitude but opposite in sign. Suppose the first term of the A.P. is 11. Find the second term.

Answer 1

Given :

• Sum of first 5 terms and first 10 terms of an A.P are equal in magnitude but opposite in sign.
• First term of an A.P is 11.

To Find :

• Find Second term of A.P

Explanation :

• It's given that first 5 terms and first 10 terms of an A.P are equal in magnitude but opposite in sign. hence,

$\boxed{\sf{S_{(5)}=-S_{(10)}}}$

• Then we can find the value of second term of the A.P by equating the above 2 values in the formula of sum of n-th term of A.P.

Formula Applied :

$\displaystyle{\underline{\boxed{\sf{S_{n}=\frac{n}{2}\[\left(2a+(n-1)d\right)\]}}}}$

$\displaystyle{\underline{\boxed{\sf{t_n = a+(n-1)d}}}}$

• a = First term of an A.P
• d = Common difference
• n = number of terms
• Sn = Sum of n-th terms of an A.P
• tn = n-th term of an A.P

Solution :

• As by explanation equate the values of $\sf{S_{(5)}}$ and $\sf{-S_{(10)}}$

$\displaystyle {\sf{\implies \frac{5}{2}\[\left(2(11)+(5-1)d\right)\]= - \frac{10}{2}\[\left(2(11)+(10-1)d\right)\]}}$

$\displaystyle {\sf{\implies \frac{5}{\cancel{2}}\[\left(22+4d\right)\]= -\frac{10}{\cancel{2}}\[\left(22+9d\right)\]}}$

$\displaystyle {\sf{\implies 5\[\left(22+4d\right)\]= -10\[\left(22+9d\right)\]}}$

$\sf{\implies 110+20d = -220-90d}$

$\sf{\implies 90d +20d = -220-110}$

$\sf{\implies 110d = -330}$

$\displaystyle\sf{\implies d = \frac{-330}{110}}$

$\sf{\boxed{ \sf d = -3}}$

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• Now substitute the values in the below formula to find the second term of the A.P

$\displaystyle{\underline{\boxed{\sf{t_n = a+(n-1)d}}}}$

$\sf{\implies t _{(2)}= 11+(2-1)(-3)}$

$\sf{\implies t _{(2)}= 11+(1)(-3)}$

$\sf{\implies t _{(2)}= 11+(-3)}$

$\sf{\boxed{\sf t _{(2)}= 8}}$

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Verification :

• As by given the sum of the first 5 and first 10 terms respectively of an A.P. Are equal in magnitude but opposite in sign.

$\boxed{\sf{S_{(5)}=-S_{(10)}}}$

$\displaystyle {\sf{\implies \frac{5}{2}\[\left(2(11)+(5-1)(-3)\right)\]= \frac{10}{2}\[\left(2(11)+(10-1)(-3)\right)}}$

$\displaystyle {\sf{\implies \frac{5}{2}\[\left(2(11)+(4)(-3)\right)\]= \frac{10}{2}\[\left(2(11)+(9)(-3)\right)}}$

$\displaystyle {\sf{\implies \frac{5}{2}\[\left(22+((-12)\right)\]= \frac{10}{2}\[\left(22+(-27)\right)}}$

$\displaystyle {\sf{\implies \frac{5}{2}\[\left(10\right)\]= \frac{10}{2}\[\left(-5\right)}}$

$\displaystyle {\sf{\implies \frac{50}{2}=\frac{-50}{2}}}$

$\textsf{Hence Verified }$

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Required Answer :

Second term of the given A.P is $\underline {\sf{-3}}$