Question

Find the sum of three consecutive terms of the A.P. whose 4th term is-15,and 9th term is -30

Answer 1

$\large\bf\underline{Given:-}$

• 4th term = -15
• 9th term = -30

$\large\bf\underline {To \: find:-}$

• Sum of three consecutive terms.

$\huge\bf\underline{Solution:-}$

Let the three consecutive terms be a - d , a and a + d.

• 4th term = -15
• 9th term = -30

$\star \bf \large \: a_n = a + (n - 1)d$

4th term = a + 3d

• ≫ a + 3d = - 15 ....(1)

9th term = a + 8d

• ≫ a + 8d = -30.....(2)

From 1) and 2)

$\tt a + 3d = -15\\ \tt a + 8d = - 30 \\ \rm \underline{(-) \: \: ( - ) \: \: \: \: \: \: ( + ) \: \: } \\ \tt \underline{ \: \: \: \: \: - 5d = \: \: \: 15}$

• d = -3

Substituting value of d in 1)

↣a + 3 × (-3) = -15

↣a -9 = -15

• ↣ a = -6

So,

three consecutive terms are :-

a - d = -6- (-3) = -3

a = -6

a + d = -6 -3 = 9

$\rule{200}3$

Answer 2

★ Question

Find the sum of three consecutive terms of the A.P. whose 4th term is-15,and 9th term is -30

★ Given

• 4th term = -15
• 9th term = -30

★ To Find

• Sum of 3 term consecutive tem = ?

★ Solution

• Let 3 consecutive term are a-d , a , a+d

$\sf→ a_4=-15\\\sf→ a-3d=-15---equ(1)$

Similarly,

$\sf→ a_9=-30\\\sf→ a-8d=-30-----equ(2)$

Subtracting equ(1) from (2)

$\sf→ a-8d=-30\\\sf→{\underline{_-a_+-3d_+=-_+30}}$

$\sf→ -5d= 15\\\sf→ {\pink{\fbox{\underline{d=-3}}}}$

★ Now,

Finding 1st term .

→ by equ(1)

$\sf→ a-3d=-15\\\sf→ a-3×3=-15\\\sf→ a= -15+3×3\\\sf→{\underline{\red{\fbox{a =-6}}}}$

Hence,

• 1st term = a - d = -6-(-3)= -3
• 2nd term = a = 9
• 3rd term = a+d -6+(-3) , -6-3 = -9