Question

In an ap consists of 37 terms the sum of three middle terms is 225 and the sum of the last three term is 429 find ap

Answer 1

Solution→

Given -

• Number of terms = 37
• Sum of middle three terms = 225
• Sum of last three terms = 429

Now , middle three terms of AP will be -

18th, 19th and 20th

Last three terms of AP -

35th , 36th and 37th

Now are given their sum so .

1st equation -

→18th +19th+20 th = 225

We can write it in another way .For example to write 2nd term of AP in another way we write it as 2nd term = a+d , Similarly

→a+17d +a+18d +a +19d = 225

→3a +54 d = 225

2nd equation -

→35 +36 +37 = 429

→ a+34d+a+35d+a+36d = 429

→ 3a +105d = 429

Now Substracting 1st equation from 2nd -

3a+105d -3a -54d = 429 -225

51 d = 204

$d = \frac{204}{51}$

d = 4

Now putting Value of d in equation 1st

3a +54(4) = 225

3a + 216 = 225

3a. = 9

a. = 3

Now we have both a and d so ap will be -

1st = a = 3

2nd = a+d = 3+4 = 7

3rd = a+2d = 3+8 = 11

and so on

AP = 3 , 7 , 11, 15 .....

Answer 2

SOLUTION:-

Given:

In an A.P. consists of 37 terms the sum of three middle terms is 225 & the sum of the last three term is 429.

To find:

The A.P.

Explanation:

Let the first term & the common difference of the A.P. are a & b respectively.

The A.P. contains 37 terms.

The middle term of the A.P. is;

$= > \frac{37 + 1}{2} th \: term \\ \\ = > \frac{38}{2} th \: term \\ \\ = > 19th \: term$

So,

Three middle term of this A.P. is, 18th ,19th & 20th term.

Given:

${}^{a} 18 + {}^{a} 19 + {}^{a} 20 = 225$

Using formula:

a+ (n-1)d

=) a+(18-1)d

=) (a+17d) + (a+18d)+(a+19d)=225

=) 3(a+18d) = 225

=) a+18d= 75

=) a= 75 -18d.............(1)

According to the question:

${}^{a} 35 + {}^{a} 36 + {}^{a} 37 = 429$

=) (a+34d) +(a+35d)+(a+36d)=429

=) 3a +105d = 429

=) 3(a+35d)= 429

=) a+35d = 143

=) a= 143 -35d

=) 75 -18d = 143 -35d [using eq.(1)]

=) -18d +35d= 143 -75

=) 17d = 68

=) d= 68/17

=) d= 4

Putting the value of equation (1), we get;

=) a= 75 - 18(4)

=) a= 75 - 72

=) a= 3

Thus,

The A.P. is 3, 7, 11,15.......

Thank you.