The 8th term of the A.P. is 0. Prove that the 38th term is triple of 18th term.
Answer
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Hint: To solve this question, we will first assume that the first term of the given AP or the arithmetic progression is ‘a’ and the common difference is ‘d’. Then we will find the 8th term and equate it equal to zero to find the value of ‘a’ in terms of ‘d’. Then we will find the 38th and 18th term and determine the relationship between the obtained values.
Complete step by step answer:
Let the first term of the given AP or the arithmetic progression is ‘a’ and the common difference is ‘d’.
It is known to us that the nth term of an AP is given by;
${a_n} = a + \left( {n - 1} \right)d$
We will put n=8 in the above formula to find the 8th term.
$
{a_8} = a + \left( {8 - 1} \right)d \\
= a + 7d \\
$
As given in the question, the 8th term of the A.P. is 0. Thus, we equate the obtained expression to 0 as follows;
$
a + 7d = 0 \\
a = - 7d \\
$
Next, we will put n=38 in the nth term formula to find the 38th term.
$
{a_{38}} = a + \left( {38 - 1} \right)d \\
= a + 37d \\
$
Substitute a=-7d into the above formula;
$
\Rightarrow {a_{38}} = - 7d + 37d \\
\Rightarrow {a_{38}} = 30d \\
$ ……(1)
Next, we will put n=18 in the nth term formula to find the 18th term.
$
{a_{18}} = a + \left( {18 - 1} \right)d \\
= a + 17d \\
$
Substitute a=-7d into the above formula;
$
\Rightarrow {a_{18}} = - 7d + 17d \\
\Rightarrow {a_{18}} = 10d \\
$ ……(2)
Thus, by (1) and (2), it can be seen that;
$
{a_{38}} = 3\left( {10d} \right) \\
= 3{a_{18}} \\
$
Hence proved that the 38th term is triple of 18th term.
Note: Whenever you face such types of problems obtain the relation between the first term and the common difference from the given conditions. Here, in this question, we have applied the formula for the nth term of the AP and found the required terms in order to use the given condition. Do remember the general formula ${a_n} = a + \left( {n - 1} \right)d$. As we have done the substitution method to find the values of common difference and first term, we can use the elimination method also to determine the same.
Complete step by step answer:
Let the first term of the given AP or the arithmetic progression is ‘a’ and the common difference is ‘d’.
It is known to us that the nth term of an AP is given by;
${a_n} = a + \left( {n - 1} \right)d$
We will put n=8 in the above formula to find the 8th term.
$
{a_8} = a + \left( {8 - 1} \right)d \\
= a + 7d \\
$
As given in the question, the 8th term of the A.P. is 0. Thus, we equate the obtained expression to 0 as follows;
$
a + 7d = 0 \\
a = - 7d \\
$
Next, we will put n=38 in the nth term formula to find the 38th term.
$
{a_{38}} = a + \left( {38 - 1} \right)d \\
= a + 37d \\
$
Substitute a=-7d into the above formula;
$
\Rightarrow {a_{38}} = - 7d + 37d \\
\Rightarrow {a_{38}} = 30d \\
$ ……(1)
Next, we will put n=18 in the nth term formula to find the 18th term.
$
{a_{18}} = a + \left( {18 - 1} \right)d \\
= a + 17d \\
$
Substitute a=-7d into the above formula;
$
\Rightarrow {a_{18}} = - 7d + 17d \\
\Rightarrow {a_{18}} = 10d \\
$ ……(2)
Thus, by (1) and (2), it can be seen that;
$
{a_{38}} = 3\left( {10d} \right) \\
= 3{a_{18}} \\
$
Hence proved that the 38th term is triple of 18th term.
Note: Whenever you face such types of problems obtain the relation between the first term and the common difference from the given conditions. Here, in this question, we have applied the formula for the nth term of the AP and found the required terms in order to use the given condition. Do remember the general formula ${a_n} = a + \left( {n - 1} \right)d$. As we have done the substitution method to find the values of common difference and first term, we can use the elimination method also to determine the same.
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