Question

in the arithmetic sequence 6,11,16 find
(a) first term (b)common difference
(c) sum of first 15 terms ​

Answer 1

Answer:

• (a) 6
• (b) 5
• (c) 615

Explanation:

Given A. P. :- 6, 11, 16

(a) first term is 6

(b) common difference is given by,

= difference first and second term

= 11 - 6

= 5

∴ Common difference is -5

(c) Sum of 15 terms

Sum of n terms of an A. P. is given by,

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

Where,

• n(number of terms) = 15
• a(first term) = 6
• d(common difference) = 5

Substituting the values,

$\\ \implies \: \rm \: S_{15} = \dfrac{15}{2} \{2(6) + (15 - 1)(5) \}$

$\implies \: \rm \: S_{15} = \dfrac{15}{2} \{ 12 + 14(5)\}$

$\implies \: \rm \: S_{15} = \dfrac{15}{2} \{ 12 + 70\}$

$\implies \: \rm \: S_{15} = \dfrac{15}{2} \times 82$

$\implies \: \rm \: S_{15} =615$

Hence, The sum of 15 terms is 615

_____________________

Formula used :

Sum of n terms of an A. P. -

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