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Q.
Write the formula for the sum of n terms in AP: ____.
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Detailed Solution
Here we need to write the formula for the sum of terms in an arithmetic progression. Let us suppose that the first term of an arithmetic progression is 'a' and the common difference is 'd'. Then our arithmetic progression will look like this,
a, a+d,a+2d,…… .
If we suppose that, this AP has n terms, then we know that, term of an AP is given by, an = a + (n−1)d so our AP becomes
a,a+d,a+2d,……,a+(n−2)d, a+(n−1)d .
Let us suppose that the last term of an AP is denoted by l i.e. a + (n − 1)d =
so the second last term becomes -d and so on. Our series looks like this,
a, a + d, a+2d,……,( −d),
Let us denote its sum by Sn so,
Sn =a+a+d+a+2d+……+ −d+ ⋯⋯⋯(1)
Similarly, if we reverse the series, the sum will remain the same, so,
Sn= + −d+ −2d+……+ a + d + a⋯⋯⋯(2)
Let us add equation (1) and (2) we get,
Sn +Sn = (a+)+(a + d+ −d)+(a+2d+ −2d)+……+( +a)
Simplifying we get,
2Sn=(a+)+(a+)+(a+)+……+(a+)n times
So adding same terms n times we get,
2Sn=n(a+)
Dividing both sides by 2, we get,
Sn = (a+).
Putting the value of as a+(n−1)d in the above formula we get,
Sn= (a+a+(n−1)d) .
Simplifying we get,
Sn = (2a + (n−1)d).
This is the required formula for the sum of n terms in an arithmetic progression.
a, a+d,a+2d,…… .
If we suppose that, this AP has n terms, then we know that, term of an AP is given by, an = a + (n−1)d so our AP becomes
a,a+d,a+2d,……,a+(n−2)d, a+(n−1)d .
Let us suppose that the last term of an AP is denoted by l i.e. a + (n − 1)d =
so the second last term becomes -d and so on. Our series looks like this,
a, a + d, a+2d,……,( −d),
Let us denote its sum by Sn so,
Sn =a+a+d+a+2d+……+ −d+ ⋯⋯⋯(1)
Similarly, if we reverse the series, the sum will remain the same, so,
Sn= + −d+ −2d+……+ a + d + a⋯⋯⋯(2)
Let us add equation (1) and (2) we get,
Sn +Sn = (a+)+(a + d+ −d)+(a+2d+ −2d)+……+( +a)
Simplifying we get,
2Sn=(a+)+(a+)+(a+)+……+(a+)n times
So adding same terms n times we get,
2Sn=n(a+)
Dividing both sides by 2, we get,
Sn = (a+).
Putting the value of as a+(n−1)d in the above formula we get,
Sn= (a+a+(n−1)d) .
Simplifying we get,
Sn = (2a + (n−1)d).
This is the required formula for the sum of n terms in an arithmetic progression.
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