Question

Write any one arithmetic progression with common difference 5. Find its nth term and sum of first 'n' terms.​

Answer 1

• The nth term and sum of first 'n' terms is (5n-3)  and (5n2-n)/2 respectively.

Step-wise solution:

Given,

• An AP having a unique quality 5.

• a common difference 5.

• To find,

• The AP's nth phrase and the sum of the first n terms have a common difference of 5.

• 2,7,12,17, is the AP with a common difference of 5.

• The total of the first n terms will be (5n2-n)/2, while the nth term will be (5n-3)

• We can simply address this problem if we follow the procedures provided.
• The common difference is the difference between two successive terms of an AP, as we know.
• The letter d stands for it Second term vs. first term is a common distinction.
• The average difference now has to be 5.
• Let's say the first term is 2, the second term is (2+5), the third term is (2+5+5), and so on.

So,

AP = 2,7,12,17,----

1st-term (a) = 2

d = 5

• We already know the formula for calculating the nth term:

an = a+(n-1)d

an = 2+(n-1)5

an = 2+5n-5

an = 5n-3

• The following is the formula for calculating the sum of the first n terms:

Sn = n/2 (a+an)

Sn = n/2 (2+5n-3)

Sn = n/2 (5n-1)

Sn = (5n²-n)/2

• As a result, with a common difference of 5, the AP is 2,7,12,17,——. (5n-3) denotes the nth term, while (5n2-n)/2 denotes the sum of the first n terms.

Answer 2

Arithmetic Progression (AP) is a sequence of numbers in order, in which the difference between any two consecutive numbers is a constant value. It is also called Arithmetic Sequence.

let the n terms be 60.

d=5

So,

The series is $5,10,15,20,...$

a=5

$a_n=a+(n-1)d\\=5+(60-1)\times5\\=5+59\times5\\=5+295\\=300$

Sum of n terms is

$S_n=\frac{n}{2} [2a+(n-1)d]\\=\frac{60}{2} [2\times5+(60-1)\times5]\\=30[10+295]\\=30\times305\\=9150$

So, the nth terms and sum of first n terms are 300 and 9150.