
If \[{a_1},{a_2},{a_3}\].....Are in A.P. and \[{a_1}^2 - {a_2}^2 + {a_3}^2 - {a_4}^2 + ...........{a_{(2k - 1)}}^2 - {a_{2k}}^2 = M({a_1}^2 - {a_{2k}}^2)\]. Then M=
A) \[\dfrac{{k - 1}}{{k + 1}}\]
B) \[\dfrac{k}{{2k - 1}}\]
C) \[\dfrac{{k + 1}}{{2k + 1}}\]
D) None
Answer
162.6k+ views
Hint: in this question we have to find the value of M present in given equation. First find the common difference and then by using formula of Sum of AP find the sum of left hand side of given equation after that compare this with right hand side expression.
Formula Used: Sum of n terms of AP in terms of first and last term is given as
\[{S_n} = \dfrac{n}{2}(a + l)\]
Where
\[{S_n}\]Sum of n terms of AP
n is number of terms
a is first term
l is last term
\[{a^2} - {b^2} = (a - b)(a + b)\]
Where
a and b are any non zero number
Complete step by step solution: Given: \[{a_1}^2 - {a_2}^2 + {a_3}^2 - {a_4}^2 + ...........{a_{(2k - 1)}}^2 - {a_{2k}}^2 = M({a_1}^2 - {a_{2k}}^2)\]
\[{a_1},{a_2},{a_3}\]……. Are in AP
So,
\[{a_2} - {a_1} = {a_3} - {a_2}\]=………………………………………..\[{a_{2k}} - {a_{2k - 1}} = d\]
Where d is common difference of AP
Now we know that \[{a^2} - {b^2} = (a - b)(a + b)\]
\[{a_1}^2 - {a_2}^2 = ({a_1} + {a_2})({a_1} - {a_2}) = - d({a_1} + {a_2})\]
\[{a_3}^2 - {a_4}^2 = ({a_3} + {a_4})({a_3} - {a_4}) = - d({a_3} + {a_4})\]
\[{a_{2k - 1}}^2 - {a_{2k}}^2 = ({a_{2k - 1}} + {a_{2k}})({a_{2k - 1}} - {a_{2k}}) = - d({a_{2k - 1}} + {a_{2k}})\]
Add all these value
\[{a_1}^2 - {a_2}^2 + {a_3}^2 - {a_4}^2 + ...........{a_{(2k - 1)}}^2 - {a_{2k}}^2 = - d({a_1} + {a_2} + {a_3}................ + {a_{2k - 1}} + {a_{2k}})\]
\[ = - d\dfrac{{2k}}{2}({a_1} + {a_{2k}}) = - dk({a_1} + {a_{2k}})\]
We know that
\[{a_{2k}} = {a_1} + (2k - 1)d\]
\[ - d = \dfrac{{{a_1} - {a_{2k}}}}{{2k - 1}}\]
Required value =\[ = \dfrac{k}{{2k - 1}}({a_1}^2 - {a_{2k}}^2)\]
\[M = \dfrac{k}{{2k - 1}}\]
Option ‘B’ is correct
Note: Here in this question it was given that some terms are in AP. So in order to find sum of AP either common difference, first term is known or first, last term is known.
If in question last term is given then there is no need to find common difference or relation between first term and common difference in order to find sum of n terms of AP
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.
Formula Used: Sum of n terms of AP in terms of first and last term is given as
\[{S_n} = \dfrac{n}{2}(a + l)\]
Where
\[{S_n}\]Sum of n terms of AP
n is number of terms
a is first term
l is last term
\[{a^2} - {b^2} = (a - b)(a + b)\]
Where
a and b are any non zero number
Complete step by step solution: Given: \[{a_1}^2 - {a_2}^2 + {a_3}^2 - {a_4}^2 + ...........{a_{(2k - 1)}}^2 - {a_{2k}}^2 = M({a_1}^2 - {a_{2k}}^2)\]
\[{a_1},{a_2},{a_3}\]……. Are in AP
So,
\[{a_2} - {a_1} = {a_3} - {a_2}\]=………………………………………..\[{a_{2k}} - {a_{2k - 1}} = d\]
Where d is common difference of AP
Now we know that \[{a^2} - {b^2} = (a - b)(a + b)\]
\[{a_1}^2 - {a_2}^2 = ({a_1} + {a_2})({a_1} - {a_2}) = - d({a_1} + {a_2})\]
\[{a_3}^2 - {a_4}^2 = ({a_3} + {a_4})({a_3} - {a_4}) = - d({a_3} + {a_4})\]
\[{a_{2k - 1}}^2 - {a_{2k}}^2 = ({a_{2k - 1}} + {a_{2k}})({a_{2k - 1}} - {a_{2k}}) = - d({a_{2k - 1}} + {a_{2k}})\]
Add all these value
\[{a_1}^2 - {a_2}^2 + {a_3}^2 - {a_4}^2 + ...........{a_{(2k - 1)}}^2 - {a_{2k}}^2 = - d({a_1} + {a_2} + {a_3}................ + {a_{2k - 1}} + {a_{2k}})\]
\[ = - d\dfrac{{2k}}{2}({a_1} + {a_{2k}}) = - dk({a_1} + {a_{2k}})\]
We know that
\[{a_{2k}} = {a_1} + (2k - 1)d\]
\[ - d = \dfrac{{{a_1} - {a_{2k}}}}{{2k - 1}}\]
Required value =\[ = \dfrac{k}{{2k - 1}}({a_1}^2 - {a_{2k}}^2)\]
\[M = \dfrac{k}{{2k - 1}}\]
Option ‘B’ is correct
Note: Here in this question it was given that some terms are in AP. So in order to find sum of AP either common difference, first term is known or first, last term is known.
If in question last term is given then there is no need to find common difference or relation between first term and common difference in order to find sum of n terms of AP
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.
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