Question

Section D 12. An A.P consists-of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 327. Find the A.P. 13 The difference of two natural numbers is 5 and the difference of their reciprocals is 1/10. Find AP


it's Isha malviya

thnx dear​

Answer 1

Hi Isha Di ...

Here is your answer

and congratulation for 10k thanks

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

-----------------------------------x----------------------------------

Answer 2

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….