Answer :

Given: AP 3,8, 13,…,253

To find: the 20^{th} term from the last term of the AP

Explanation: The given AP is 3,8, 13,….,253

So, the first term is 3

∴ a = 3……(i)

Now finding the difference of second and term, we get

d = 8 - 3 = 5………(ii)

So last term can be written as

a_{n} = 253

But we know a_{n} = a + (n - 1)d (standard form)

⇒ 253 = a + (n - 1)d

Substituting the values from equation (i) and (ii), we get

⇒ 253 = 3 + (n - 1)(5)

⇒ 253 = 3 + 5n - 5

⇒ 253 = 5n - 2

⇒ 5n = 253 + 2

⇒ 5n = 255

⇒ n = 51

Hence 253 is the 51^{st} term of the given AP

Now to the 20^{th} term from last means 51 - 19 = 32^{nd} term, as the last term is 51 - 0, second last term is 51 - 1, third last term is 51 - 2 and so on.

So 32^{nd} term can be found from using the standard form formula, i.e., a_{n} = a + (n - 1)d, so

a_{32} = a + (32 - 1) d

Substituting values of a and d from equation (i) and (ii), we get

a_{32} = 3 + (31)(5)

⇒ a_{32} = 3 + 155

⇒ a_{32} = 158

Hence the 20^{th} term from the last term of the AP is 158.

**OR**

Given: 7 times the 7^{th} term of an A.P is equal to 11 times its 11^{th} term

To find: its 18^{th} term

Explanation: The standard form of an AP term is

a_{n} = a + (n - 1) d

Now the 7^{th} term is,

a_{7} = a + (7 - 1) d = a + 6d……..(i)

And the 11^{th} term is,

a_{11} = a + (11 - 1)d = a + 10d……..(ii)

And also the 18^{th} term is,

a_{18} = a + (18 - 1)d = a + 17d……..(iii)

Now as per the given criteria,

7 times the 7^{th} term of an A.P is equal to 11 times its 11^{th} term

⇒ 7 × a_{7} = 11 × a_{11}

Now substituting values form equation (i) and (ii), we get

⇒ 7 × (a + 6d) = 11 × (a + 10d)

⇒ 7a + 42d = 11a + 110d

⇒ 7a - 11a = 110d - 42d

⇒ - 4a = 68d

⇒a = - 17d…….(iv)

Now substituting equation (iv) in equation (iii), we get

a_{18} = a + 17d

⇒ a_{18} = (-17d) + 17d = 0

Hence the 18^{th} term is 0.

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