Question

an ap consists of 37 terms. the sum of the three middle most terms is 225 and the sum of the last three terms is 429. find the AP

Answer 1

Let the first term and the common difference of the A.P are a and d respectively.

Since the A.P contains 37 terms. So, the middle most term is (37+1)/2 th term = 19th term.

Thus, three middle most terms of this A.P.are 18th, 19th and 20th terms.

Given a 18 + a 19 + a 20 = 225

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

⇒ 3(a + 18d) = 225

⇒ a + 18d = 75

⇒ a = 75 – 18d … (1)

 

According to given information

a 35 + a 36 + a 37 = 429

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

⇒ 3(a + 35d) = 429

⇒ (75 – 18d) + 35d = 143

⇒ 17d = 143 – 75 = 68

⇒ d = 4

 

Substituting the value of d in equation (1), it is obtained

a = 75 – 18 × 4 = 3

 

Thus, the A.P. is 3, 7, 11, 15

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

${\boxed{\sf\:{First\;term\;be\;p}}}$

Also

${\boxed{\sf\:{Common\;difference\;be\;d}}}$

Given,

Middle term is 19th

n = 37

Three middle most terms are 18th , 19th and 20th

$\tt{\rightarrow p_{18}+p_{19}+p_{20}=225}$

p + 17d + p + 18d + p + 19d = 225

3p + 54d = 225

3(p + 18d) = 225

$\tt{\rightarrow p+18d=\dfrac{225}{3}}$

p + 18d = 75 ..... (1)

Also we have

$\tt{\rightarrow p_{35}+p_{36}+p_{37}=429}$

p + 34d + p + 35d + p + 36d = 429

3p + d(34 + 35 + 36) = 429

3p + 105d = 429

3(p + 35d) = 429

$\tt{\rightarrow p+35d=\dfrac{429}{3}}$

p + 35d = 143 ...... (2)

Subtracting (1) from (2)

17d = 68

$\tt{\rightarrow d=\dfrac{68}{17}}$

d = 4

Put the value of d in (2)

p + 35d = 143

p + 35(4) = 143

p + 140 = 143

p = 143 - 140

p = 3

$\Large{\boxed{\sf\:{AP=3,7,11,15,19}}}$