Question

in a AP 1,7,13,19....415 prove that the sum of the middle terms is equal to the firsr and last terms

Answer 1

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Answer 2

Proved below.

Step-by-step explanation:

Given:

In a AP 1, 7, 13, 19, ....415, we have

a = 1, d = second term - first term = 7 -1 = 6 and last term = l =415.

So,

l = a+(n−1)d

⇒ 415 = 1+(n−1)6

⇒ 415 -1 = (n−1)6

⇒ 414 = (n−1)6

⇒ 414 = 6n − 6

⇒ 414 + 6 = 6n

⇒ 420 = 6n

⇒ $n = \frac{420}{6}$

⇒n=70

Clearly its middle terms are 35th and 36th,

$a_{35} = a+(35-1)d = 1 + (34\times6) = 205$

$a_{36}=a+(36-1)d=1+(35\times6)=211$

Sum of middle terms = $a_{35}+a_{36}$ = 205+211 =416       [1]

And sum of first and last term = a + l = 1 + 415 = 416                             [2]

From eq 1 and 2,

Sum of the middle terms is equal to the sum of first and last terms .

Hence proved.