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 the second and first term, we get

d = 8 – 3 = 5………(ii)

So the 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_{7 }= a + 6d……..(i)

And the 11^{th} term is,

a_{11} = a + (11 – 1)d

a_{11 }= a + 10d……..(ii)

And also the 18^{th} term is,

a_{18} = a + (18 – 1)d

a_{18 }= 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 from 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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