The sum of the third term and seventh term of an AP is 6 and their product is 8. Find the first term and the common difference of the AP.
Answer
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Hint – In order to solve this problem, make the equations according to the question using the formula of general term of an AP and solve the equation to get the value asked.
The general term of an AP is,
${{\text{a}}_{\text{n}}}{\text{ = a + (n - 1)d}}$
Here we know from the question that ,
${{\text{a}}_{\text{3}}}{\text{ + }}{{\text{a}}_{\text{7}}}{\text{ = 6}}$
Expanding using the formula of general term in above equation as,
$
{\text{a + (3 - 1)d + a + (7 - 1)d = 6}} \\
{\text{2a + 8d = 6}} \\
$
${\text{a + 4d = 3}}$ ……(i)
And,
${{\text{a}}_{\text{3}}}{{\text{a}}_{\text{7}}}{\text{ = 8}}$
On expanding above equation we get,
${\text{(a + 2d)(a + 6d) = 8}}$
${{\text{a}}^{\text{2}}}\,{\text{ + }}\,{\text{12}}{{\text{d}}^{\text{2}}}\,{\text{ + }}\,{\text{8ad = 8}}$ ……(ii)
From (i) we can say ,
${\text{a = 3 - 4d}}$
On putting the value of a in equation (ii) we get,
$
{{\text{(3 - 4d)}}^{\text{2}}}{\text{ + 8(3 - 4d)d + 12}}{{\text{d}}^{\text{2}}}{\text{ = 8}} \\
{\text{9 + 16}}{{\text{d}}^{\text{2}}}{\text{ - 24d + 24d - 32}}{{\text{d}}^{\text{2}}}{\text{ + 12}}{{\text{d}}^{\text{2}}}{\text{ = 8}} \\
{\text{1 - 4}}{{\text{d}}^{\text{2}}}{\text{ = 0}} \\
{\text{(2d + 1)(2d - 1) = 0}} \\
$
Either 2d = -1 or 2d = 1
In above equations we will use the following formulas,
$
{{\text{(a - b)}}^2} = {{\text{a}}^2} - {{\text{b}}^2}{\text{ - 2ab}} \\
{\text{& }} \\
{{\text{a}}^2} - {{\text{b}}^2} = ({\text{a - b}})({\text{a + b}}) \\
$
Therefore we can say,
$
{\text{d = }}\dfrac{{{\text{ - 1}}}}{{\text{2}}} \\
{\text{or}} \\
{\text{d = }}\dfrac{{\text{1}}}{{\text{2}}} \\
$
Therefore the common difference of this series can be $\dfrac{{{\text{ - 1}}}}{{\text{2}}}\,\,\,{\text{or }}\,\dfrac{{\text{1}}}{{\text{2}}}$.
Now calculation of first term can be done as ,
When d = $\dfrac{{{\text{ - 1}}}}{{\text{2}}}\,\,$
Then on putting the value of d in equation (i) we get,
a + 4($\dfrac{{{\text{ - 1}}}}{{\text{2}}}\,\,$) = 3
Then we do,
a – 2 = 3
Therefore a = 5.
And,
When d = $\dfrac{{\text{1}}}{{\text{2}}}\,\,$
Then on putting the value of d in equation (i) we get,
a + 4($\dfrac{{\text{1}}}{{\text{2}}}\,\,$) = 3
Then we do,
a + 2 = 3
Therefore a = 1
Hence there is Two Arithmetic progressions,
The first AP’s first term is 5 and the common difference is $\frac{{{\text{ - 1}}}}{{\text{2}}}\,\,$ .
The second AP’s first term is 1 and the common difference is $\dfrac{{\text{1}}}{{\text{2}}}\,\,$.
Note – In these types of questions we have to just obtain the equation from the question provided, then solve those equations to get the value of variables asked. Here we have used the formula of the general term of an AP to expand the equations. Then we have solved the equation to get the value of common difference and first term.
The general term of an AP is,
${{\text{a}}_{\text{n}}}{\text{ = a + (n - 1)d}}$
Here we know from the question that ,
${{\text{a}}_{\text{3}}}{\text{ + }}{{\text{a}}_{\text{7}}}{\text{ = 6}}$
Expanding using the formula of general term in above equation as,
$
{\text{a + (3 - 1)d + a + (7 - 1)d = 6}} \\
{\text{2a + 8d = 6}} \\
$
${\text{a + 4d = 3}}$ ……(i)
And,
${{\text{a}}_{\text{3}}}{{\text{a}}_{\text{7}}}{\text{ = 8}}$
On expanding above equation we get,
${\text{(a + 2d)(a + 6d) = 8}}$
${{\text{a}}^{\text{2}}}\,{\text{ + }}\,{\text{12}}{{\text{d}}^{\text{2}}}\,{\text{ + }}\,{\text{8ad = 8}}$ ……(ii)
From (i) we can say ,
${\text{a = 3 - 4d}}$
On putting the value of a in equation (ii) we get,
$
{{\text{(3 - 4d)}}^{\text{2}}}{\text{ + 8(3 - 4d)d + 12}}{{\text{d}}^{\text{2}}}{\text{ = 8}} \\
{\text{9 + 16}}{{\text{d}}^{\text{2}}}{\text{ - 24d + 24d - 32}}{{\text{d}}^{\text{2}}}{\text{ + 12}}{{\text{d}}^{\text{2}}}{\text{ = 8}} \\
{\text{1 - 4}}{{\text{d}}^{\text{2}}}{\text{ = 0}} \\
{\text{(2d + 1)(2d - 1) = 0}} \\
$
Either 2d = -1 or 2d = 1
In above equations we will use the following formulas,
$
{{\text{(a - b)}}^2} = {{\text{a}}^2} - {{\text{b}}^2}{\text{ - 2ab}} \\
{\text{& }} \\
{{\text{a}}^2} - {{\text{b}}^2} = ({\text{a - b}})({\text{a + b}}) \\
$
Therefore we can say,
$
{\text{d = }}\dfrac{{{\text{ - 1}}}}{{\text{2}}} \\
{\text{or}} \\
{\text{d = }}\dfrac{{\text{1}}}{{\text{2}}} \\
$
Therefore the common difference of this series can be $\dfrac{{{\text{ - 1}}}}{{\text{2}}}\,\,\,{\text{or }}\,\dfrac{{\text{1}}}{{\text{2}}}$.
Now calculation of first term can be done as ,
When d = $\dfrac{{{\text{ - 1}}}}{{\text{2}}}\,\,$
Then on putting the value of d in equation (i) we get,
a + 4($\dfrac{{{\text{ - 1}}}}{{\text{2}}}\,\,$) = 3
Then we do,
a – 2 = 3
Therefore a = 5.
And,
When d = $\dfrac{{\text{1}}}{{\text{2}}}\,\,$
Then on putting the value of d in equation (i) we get,
a + 4($\dfrac{{\text{1}}}{{\text{2}}}\,\,$) = 3
Then we do,
a + 2 = 3
Therefore a = 1
Hence there is Two Arithmetic progressions,
The first AP’s first term is 5 and the common difference is $\frac{{{\text{ - 1}}}}{{\text{2}}}\,\,$ .
The second AP’s first term is 1 and the common difference is $\dfrac{{\text{1}}}{{\text{2}}}\,\,$.
Note – In these types of questions we have to just obtain the equation from the question provided, then solve those equations to get the value of variables asked. Here we have used the formula of the general term of an AP to expand the equations. Then we have solved the equation to get the value of common difference and first term.
Last updated date: 21st Sep 2023
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