Question

In the AP find the missing term in the boxes.
-4,box,box,box,box,6

Answer 1

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

• AP:- -4 ,❐,❐,❐,❐, 6

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

• Missing terms.

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

• First term (a) = -4
• 6th term = 6

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

$\tt \: a_6 = a + (6 - 1)d \\ \\ \tt \: a_6 = a + 5d \\ \\ \tt \: a + 5d= 6......(i)$

✝️ Substituting value of a = -4 in (i)

$\dashrightarrow\tt \: a + 5d = 6 \\ \\ \dashrightarrow\tt \: - 4 + 5d = 6 \\ \\ \dashrightarrow\tt \: 5d = 6 + 4 \\ \\ \dashrightarrow\tt \: 5d = 10 \\ \\ \dashrightarrow\tt \: d = \cancel\frac{10}{5} \\ \\ \dashrightarrow\tt \: d = 2$

Now ,

»★ 2nd term a + d = -4 + 2 = -2

»★ 3rd term a + 2d = -4+ 2×2 = 0

»★ 4th term a + 3d = -4 +3 × 2 = 2

»★ 5th term a + 4d = -4 + 4 × 2 = 4

$\boxed{ \bf - 4 \:, \boxed{ - 2 } , \: \boxed{0} \:, \boxed{2} \: ,\boxed{4},6}$

Answer 2

Answer:

$\sf{-4, \ \boxed{-2}, \ \boxed{0}, \ \boxed{2}, \ \boxed{4}, \ 6.}$

Given:

• $\sf{First \ term \ (a)=-4}$
• $\sf{6^{th} \ term \ (t_{6})=6}$

To find:

• Missing terms.
• $\sf{t_{2}, \ t_{3}, \ t_{4} \ and \ t_{5}}$

Solution:

$\boxed{\sf{t_{n}=a+(n-1)d}}$

$\sf{\therefore{t_{6}=a+(6-1)d}}$

$\sf{But, \ a=-4 \ and \ t_{6}=6}$

$\sf{\therefore{6=-4+5d}}$

$\sf{\therefore{5d=6+4}}$

$\sf{\therefore{5d=10}}$

$\sf{\therefore{d=\frac{10}{5}}}$

$\sf{\therefore{d=2}}$

________________________________

$\sf{Here, \ a=-4 \ and \ d=2}$

$\sf{t_{2}=a+d=-4+2=-2,}$

$\sf{t_{3}=a+2d=-4+2(2)=0,}$

$\sf{t_{4}=a+3d=-4+3(2)=2,}$

$\sf{t_{5}=a+4d=-4+4(2)=4.}$

$\sf\purple{\tt{-4, \ \boxed{-2}, \ \boxed{0}, \ \boxed{2}, \ \boxed{4}, \ 6.}}$