Question

find the 10th term from the least term of AP -1,5/-6,2/-3,1/-2------10/3​

Answer 1

$\red{\huge{\boxed{\sf Correct\ Question:}}}$

Find the 10th term from the last term of AP

$\implies (-1),\ \dfrac{5}{-6},\ \dfrac{2}{-3},\ \dfrac{1}{-2}........ \dfrac{10}{3}$

$\red{\huge{\boxed{\sf{Solution:}}}}$

To find nth term from the end of an AP, we can use

$\green{\boxed{\boxed{ \tt{n^{th}\ term\ from\ end\ of\ an\ AP} = l-(n-1)d}}}$

[where 'l ' is the last term, 'n' is the nth term, and 'd' is common difference]

So,

Fist term = 'a' or 'a₁' = (-1)

and,

Common difference = 'd'

→ a₂ - a₁

$= \dfrac{-5}{6} - (1)$

[taking LCM = 6]

$= \dfrac{-5-(-6)}{6} = \dfrac{-5+6}{6} = \boxed{\boxed{\dfrac{1}{6}}}$

Last term = 'l' = $\dfrac{10}{3}$

Now,

$\bf a_{10}\ term\ from\ the\ end\ of\ AP = l-(n-1)d \\ \\ = \frac{10}{3} - (10-1)(\frac{1}{6}) \\ \\ = \frac{10}{3}-(9)(\frac{1}{6}) \\ \\ = \frac{10}{3} - \frac{9}{6} \\ \\ \big[ dividing\ 9\ and\ 6\ by\ 3\big] \\ \\ = \frac{10}{3} - \frac{3}{2} \\ \\ \bigg[Taking\ LCM\ 6 \bigg]\\ \\ =\frac{20-9}{6} \\ \\ = \frac{11}{6}$

Hence,

Answer is

$\boxed{\boxed{\bf \dfrac{11}{6}}}$

Answer 2

${\underline{\sf{Question}}}$

Find the 10th term from the last term of AP $\sf{(-1),\dfrac{5}{-6},\dfrac{2}{-3}, \dfrac{1}{-2}........ \dfrac{10}{3}}$

${\underline{\sf{Theory}}}$

We know that , nth term from the end of an AP is :

${\purple{\boxed{\large{\bold{n^{th}\ term\ from\ end\ of\ an\ AP= l-(n-1)d}}}}}$

where

• 'l ' is the last term,
• 'n' is the nth term,
• and 'd' is common difference

${\underline{\sf{Solution}}}$

We have :

Fist term,$\sf\:a_1=a=-1$

Common difference ,d

$\sf\:=a_2-a_1$

$\sf\:= \frac{-5}{6} - 1$

$\sf=\frac{-5-(-6)}{6} = \frac{-5+6}{6}$

$\sf=\dfrac{1}{6}$

Last term,$\sf\:l= dfrac{10}{3}$

We know that :

$\sf{n^{th}\ term\ from\ end\ of\ an\ AP= l-(n-1)d}$

then ,$\sf{a_{10}}$term from the end ofAP

= l-(n-1)d

$\sf{= \frac{10}{3} - (10-1)\times\frac{1}{6}}$

$\sf{= \frac{10}{3}-(9)\times\frac{1}{6}}$

$\sf{= \frac{10}{3} - \frac{9}{6}}$

$\sf{=\frac{10}{3} - \frac{3}{2}=\frac{20-9}{6}}$

$\sf{=\frac{11}{6}}$

Therefore,the 10th term from the last term of given AP is $\sf{\dfrac{11}{6}}$

$\rule{200}2$

More About Arithmetic Progression:

•Genral term of an Ap

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

•Sum of first n terms of an AP

$\sf \: s_{n} = \dfrac{n}{2} (2a + (n - 1)d)$