Question

Check whether -150 is the term of AP is such that 11,8,5,2....

Answer 1

$\bold \red \star \: \underline{Given:-} \begin {cases} \sf{ AP :- \: } \\ \sf{11 ,\: 8,\: 5,\: 2,\: ........} \\ \sf{a_n = - 150} \end{cases}$

$\bold \red \star \: \underline{Solution:-}$

$\longrightarrow \: \green a =\green{ 11.} \: \: \:$

$\longrightarrow \: \blue d = a_2-a_1. \\ \\ \longrightarrow \: \: \: \: \: \: = 8 - 11. \\ \\ \longrightarrow \: \: \: \: \: \: = \blue {- 3.}$

$\: \: \: \: \: \: \: \: \: \: a_n = a + (n - 1)d. \\ \\ \: \: \: \: - 150 = 11 + (n - 1) - 3. \\ \\ \frac{\cancel{ - } \cancel{150}}{ \cancel{-} \cancel {3}} = 11 + (n - 1). \\ \\ 50 - 11 = n - 1. \\ \\ \: \: 39 + 1 = n. \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: n = \red {40.}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\:\:\: \:\:\:\:{ \huge {\underline { \boxed{n = \red{ 40.}}}}}$

$Yes, \: \red{ -150} \: \: is \: the \: term \: of \: AP.$

Answer 2

Answer:

Given -

• $\sf{a}_{n}$ = -150
• a = 11
• d = 8 - 11 = -3
• n = ?

We know that,

$\\ \implies \sf \: a_n = a + (n - 1) \: d \\ \\ \\ \implies \sf \: - 150 = 11 + (n - 1) \: (- 3) \\ \\ \\ \implies \sf \: - 150 = 11 - 3n + 3 \\ \\ \\ \implies \sf \: - 150 = 14 - 3n \\ \\ \\ \implies \sf \: - 150 - 14 = 3n \\ \\ \\ \implies \sf \: - 164 = 3n \\ \\ \\ \sf \: or, \: 3n = 164 \\ \\ \\ \implies \sf \: n = \frac{164}{3}$

which is not an integral number.

Hence, -150 is not a term of the AP.