Question

find the 20th term of the A.P 15,12,9,6.....​

Answer 1

⠀

$\huge\sf\green{\underline{\underline{Given\: :-}}}$

$\sf{}$

$\sf \large{ t_{1} = a = 15 \: } \\ \sf \large{ t_{2}\: = 12 \:} \: \: \: \: \: \: \\ \sf \large{ t_{3} \:= \: 9 \: } \: \: \: \: \: \: \: \\ \sf \large{ t_{4} = 6} \: \: \: \: \: \: \: \: \: \:$

$\sf{}$

$\huge\sf\green{\underline{\underline{To \: find \: :-}}}$

$\sf{}$

$\sf \large \: { t_{20 \:} = \: \: ?}$

$\sf{}$

$\huge\sf\green{\underline{\underline{Formula\: used \: :-}}}$

$\sf{}$

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

$\sf{}$

$\huge\sf\green{\underline{\underline{Solution \: :-}}}$

$\sf{}$

According to the question,

To find the 20th term of the A.P. ,

The value of d must be known

So, d = ?

$\sf{}$

$\sf \large{d = t_{2} \: - \: t_{1}= 12 \: - 15 = \red{ - \: 3}} \\ \sf \large{d = \: t_{3} \: - \: t_{2}= 9 \: - 12 \: = \red{ - \: 3}} \\ \sf \large{d = \: t_{4} \: - \: t_{3} \: = \: 6 \: - 9 \: = \red{ - \: 3}} \\ \\ \sf \large \purple{ \therefore \: d \: = \: - \: 3 \: \: \: \; \: \; \; \: \: \: \: \: \; \: \; \; \: \: \: \: \; \: \; \; \:\: \; \: \; \; \:\:}$

$\sf{}$

We got all the required values ,

So now we will substitute them.

$\sf{}$

$\sf \large \red{{ t_{n}} = a + (n - 1) \: d} \: \: \: \: \: \: \: \: \: \: \: \\ \sf \large{{ t_{20}} = 15 + (20- 1) \: - 3} \\ \sf \large{{ t_{20}} = 15 + 19 \times \: - 3} \: \: \: \: \: \: \: \\ \sf \large{{ t_{20}} = 15 \: - 57} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \large {{ t_{20}} = - \: 42} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf{}$

Thus the 20th term of the A.P. is -42

Answer 2

$\huge\blue{\textbf{\textsf{question:}}}$

$\bf find \: the \: {20}^{th} \: term \: of \: the \: ap : \\ \bf 15 \: \: 12 \: \: 9 \: \: and \: \: 6$

$\huge\blue{\textbf{\textsf{answer:}}}$

$\bf{Given:}$

$\sf{~~~~~~~~a=15}$

$\sf{~~~~~~~~d=a_{2}-a_{1}}$

$\sf{~~~~~~~~~~~=12-15}$

$\sf{~~~~~~~~d={\underline{\boxed{-3}}}}$

$\sf{~~~~~~~~n=20}$

$\sf{ }$

$\sf{~~ \therefore a_{n}=a+(n-1)d}$

$\sf{~~~~a_{20}=15+(20-1)-3}$

$\sf{~~~~~~~~~~=15+(19)-3}$

$\sf{~~~~~~~~~~=15-57}$

$\sf{\therefore a_{20}={\underline{\boxed{\bf -42}}}}$

$\bf{Hence,~20^{th}~term~of~the~AP~is~~-42}$