
How many terms of the A.P. 17, 15, 13, …. must be taken so that their sum is 81?
Answer
592.2k+ views
Hint: Here we can consider the number of terms required to be ‘n’ and then we may apply the formula for the sum of terms of an A.P. and can find the value of ‘n’.
Complete step by step answer:
Here, we can see that the first term in this given A.P. is 17 and the second term is 15.
Since, we know that the common difference of an A.P. is given as the difference of a term and the term just next to it.
So, here in this case also we can find the common difference of this A.P. by subtracting the 1st term from the 2nd term or also 2nd term from 3rd term and so on.
So, common difference ‘d’ for this A.P. is:
d = 15 – 17 = -2
Also we may check it by subtracting 3rd term from 2nd term which is:
13 – 15 = -2 =d
Hence, now we are sure that the common difference of this A.P. is (-2).
Now we may consider that the number of terms of this A.P. required to get a sum of 81 be = ‘n’.
So, now we can use the formula for sum of ‘n’ terms of an A.P. which is given as:
${{S}_{n}}=\dfrac{n}{2}\left\{ 2a\,+\left( n-1 \right)d \right\}..............(1)$
Since, we have, ${{S}_{n}}=81$ , a = 17 and d = ( -2 ) , so now on substituting these values in equation (1) , we get :
$81=\dfrac{n}{2}\left\{ 2\times 17\,+\left( n-1 \right)\left( -2 \right) \right\}$
Or, $81=\dfrac{n}{2}\left\{ 34-2n+2 \right\}$
Or, $81\times 2=n\left( 36-2n \right)$
Or, $162=36n-2{{n}^{2}}$
Or, $2{{n}^{2}}-36n+162=0$
Or, ${{n}^{2}}-18n+81=0$
So, now we have got a quadratic equation in n, so now we may solve it and get the value of n.
So, we can use the factorization method to solve this quadratic equation.
${{n}^{2}}-9n-9n+81=0$
Or, $n\left( n-9 \right)-9\left( n-9 \right)=0$
Or, $\left( n-9 \right)\left( n-9 \right)=0$
So, from here we get the value of n = 9.
Hence, the number of terms of the given A.P. required to give a sum of 81 is 9.
Note: Students should take care that the common difference of an A.P., which is only given by subtracting a term from its succeeding term but not by previous term, so here common difference will only be (15-17) but not (17-15).
Complete step by step answer:
Here, we can see that the first term in this given A.P. is 17 and the second term is 15.
Since, we know that the common difference of an A.P. is given as the difference of a term and the term just next to it.
So, here in this case also we can find the common difference of this A.P. by subtracting the 1st term from the 2nd term or also 2nd term from 3rd term and so on.
So, common difference ‘d’ for this A.P. is:
d = 15 – 17 = -2
Also we may check it by subtracting 3rd term from 2nd term which is:
13 – 15 = -2 =d
Hence, now we are sure that the common difference of this A.P. is (-2).
Now we may consider that the number of terms of this A.P. required to get a sum of 81 be = ‘n’.
So, now we can use the formula for sum of ‘n’ terms of an A.P. which is given as:
${{S}_{n}}=\dfrac{n}{2}\left\{ 2a\,+\left( n-1 \right)d \right\}..............(1)$
Since, we have, ${{S}_{n}}=81$ , a = 17 and d = ( -2 ) , so now on substituting these values in equation (1) , we get :
$81=\dfrac{n}{2}\left\{ 2\times 17\,+\left( n-1 \right)\left( -2 \right) \right\}$
Or, $81=\dfrac{n}{2}\left\{ 34-2n+2 \right\}$
Or, $81\times 2=n\left( 36-2n \right)$
Or, $162=36n-2{{n}^{2}}$
Or, $2{{n}^{2}}-36n+162=0$
Or, ${{n}^{2}}-18n+81=0$
So, now we have got a quadratic equation in n, so now we may solve it and get the value of n.
So, we can use the factorization method to solve this quadratic equation.
${{n}^{2}}-9n-9n+81=0$
Or, $n\left( n-9 \right)-9\left( n-9 \right)=0$
Or, $\left( n-9 \right)\left( n-9 \right)=0$
So, from here we get the value of n = 9.
Hence, the number of terms of the given A.P. required to give a sum of 81 is 9.
Note: Students should take care that the common difference of an A.P., which is only given by subtracting a term from its succeeding term but not by previous term, so here common difference will only be (15-17) but not (17-15).
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