1. Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.
2. The first and last terms of an AP are a and l respectively. Show that the sum of the nth term from the beginning and the nth term from the end is (a+l) .
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Answers
Let d is common difference of AP
Now first 4 terms are 5, 5+d, 5+2d, 5+3d
and next 4 terms 5+4d, 5+5d, 5+6d, 5+7d
Given that, the sum of its first four terms is half the sum of the next four terms.
i.e.,
5 + 5+d + 5+2d + 5+3d=25+4d + 5+5d + 5+6d + 5+7d/2
20+6d=2(20+22d)/2
20+6d=10+11d
d=2
Hence, the common difference of the given A.P. is 2
Let d is common difference of AP
Now first 4 terms are 5, 5+d, 5+2d, 5+3d
and next 4 terms 5+4d, 5+5d, 5+6d, 5+7d
Given that, the sum of its first four terms is half the sum of the next four terms.
i.e.,
5 + 5+d + 5+2d + 5+3d=
2
5+4d + 5+5d + 5+6d + 5+7d
20+6d=
2
(20+22d)
20+6d=10+11d
d=2
Hence, the common difference of the given A.P. is 2
2). a
1
=a
Let
ak=l
a+(k−1)d=l
d=
k−1
l−a
n
th
term from beginning =a
n
=a+(n−1)d=a+(n−1)(
k−1
l−a
)
n
th
term from end =l+(n−1)(−d)=l−(n−1)(
k−1
l−a
)
Sum of n
th
terms from beginning and from end =a+(n−1)(
k−1
l−a
)+l−(n−1)(
k−1
l−a
)=a+l
proved.
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