Question

The sum of n terms of an ap whose first twrm is 5 and common difference is 36

Answer 1

Solution:

Given that ;

• a = 5
• d = 36

We know that :

$\rm S_n = \dfrac{n}{2} [2a + (n - 1)d]$

$\rm S_n = \frac{n}{2}[2(5) + (n - 1) \times 36] \\ \\ \rm S_n = \frac{n}{2}[10 + 36n - 36 ] \\ \\ \rm S_n = \frac{n}{2}[36n - 26] \\ \\ \rm S_n = \frac{n(36n - 26)}{2} \\ \\ \rm S_n = \frac{36 {n}^{2} - 26n}{2} \\ \\ \rm S_n = \frac{2n(18n - 13)}{2} \\ \\ \rm S_n = n(18n - 13) \\ \\ \rm S_n = {18n}^{2} - 13n$

Hence, the sum of n terms of A.P is (18n² - 13n).

Answer 2

GIVEN:

The first term of an AP = a = 5

Common Difference = d = 36

We know that,

Sum of terms in an AP, Sn = n/2 [ 2a + (n - 1)d ]

Sn = n/2 [ 2a + (n - 1)d ]

Sn = n/2 [ 2(5) + (n - 1)36]

Sn = n/2 [ 10 + 36n - 36 ]

Sn = n/2 [ 36n - 26 ]

Sn = (36n² - 26n) / 2

Sn = 18n² - 13n

Therefore, the sum of n terms is 18n² - 13n.