Question

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) .

→ Please any one give the answer of above question with explanation...​

Answer 1

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

Answer 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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