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Hi Isha Di ...

Here is your answer

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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 …

Hope it helps

@s8a1583aritra1756

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Answer:

Hi isha di

Yumm!!

I want to eat that cake

Step-by-step explanation:

We know that,

First term of an AP = a

Common difference of AP = d

nth term of an AP, an = a + (n – 1)d

Since, n = 37 which is an odd number

Middle term will be (n+1)/2 = 19thterm

Thus, the three middle most terms will be,

18th, 19th and 20th terms

According to the question,

a18 + a19 + a20 = 225

Using an = a + (n – 1)d

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

3a + 54d = 225

3a = 225 – 54d

a = 75 – 18d … (1)

Now, we know that the last three terms will be 35th, 36th and 37th terms.

According to the question,

a35 + a36 + a37 = 429

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

3a + 105d = 429

a + 35d = 143

Substituting a = 75 – 18d from equation 1,

75 – 18d + 35d = 143 [using eqn1]

17d = 68

d = 4

Then,

a = 75 – 18(4)

a = 3

Therefore, the AP is a, a + d, a + 2d….

i.e. 3, 7, 11….