Question

Find the number of terms in the AP 3,7,11.....407 also find its 20th term from the end

Answer 1

Step-by-step explanation:

difference=a2-a1=7-3=4

an=a+(n-1)d

407=3+(n-1)4

107=3+4n-4

107-3+4=4n

108=4n

108/4=n

27=n

the 20th term from last will be 7th term

a27=3+(7-1)×4

=3+6×4

=3+24

=27

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Answer 2

Answer:

There are 102 terms in this AP and 331 is the 20th term from the last.

Step-by-step explanation:

We have,

A.P. = 3, 7, 11,........, 407

a = 3

Common difference(d) = a2 - a1

d = 7 - 3

d = 4

Thus, A.P = 3, 7, 11,......, 403, 407

We know that,

a(nth) = a + (n - 1)d

Last term (l) = 407

a(nth) = 407

a = 3

d = 4

407 = 3 + (n - 1)4

407 - 3 = (n - 1)4

404/4 = n - 1

101 = n - 1

n = 101 + 1

n = 102

Thus, the 102nd term is the last term,

Now, we must find the 20th term from the last,

so, let's take it in reverse order

Thus,

A.P = 407, 403,......, 11, 7, 3

Now,

d = 403 - 407

d = -4

[Just remember when we take the reverse order of an AP, the last term becomes the first term and the first term becomes the last term

also, the common difference will become negative if it was positive and it becomes positive if it was negative.

For ex:- Let the AP be 1, 2, 3, 4,......, 10

Now, its common difference = 2 - 1 = 1

and the first term is 1

Now, when it is reversed the AP becomes 10, 9,......, 2, 1

Now, common difference = 9 - 10 = -1

and first term = 10]

So,

AP = 407, 403, ...... , 2, 1

a = 407

d = -4

a(nth) = a + (n - 1)d

a(20) = 407 + (20 - 1)(-4)

= 407 + (19)(-4)

= 407 - 76

= 331

Thus, the 20th term from last = 331