# Find the 20t

Given: AP 3,8, 13,…,253

To find: The 20th 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

an = 253

But we know an = 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 51st term of the given AP.

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

So 32nd term can be found from using the standard form formula, i.e.,

an = a + (n – 1)d, so

a32 = a + (32 – 1) d

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

a32 = 3 + (31)(5)

a32 = 3 + 155

a32 = 158

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

OR

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

To find: its 18th term

Explanation: The standard form of an AP term is

an = a + (n – 1) d

Now the 7th term is,

a7 = a + (7 – 1) d

a= a + 6d……..(i)

And the 11th term is,

a11 = a + (11 – 1)d

a11 = a + 10d……..(ii)

And also the 18th term is,

a18 = a + (18 – 1)d

a18 = a + 17d……..(iii)

Now as per the given criteria,

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

7 × a7 = 11 × a11

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

a18 = a + 17d

a18 = (–17d) + 17d

= 0

Hence the 18th term is 0.

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