Question

If the 11th term of an arithmetic progression is 43 and the first term is 3 what is the common difference

Answer 1

⇝ Given :-

For An Arthematic Progression ;

• 11th Term =  43

• First term = 3

⇝ To Find :-

• Common Difference of the corresponding Arthematic Progression.

⇝ Solution :-

We Know, nth term of an Arthematic Progression is given by formula :

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{  a_n = a +(n - 1)d}}}$

where,

• a = First Term

• d = Common Difference

Here,

• $\rm First \:Term = a = 3$

• $\rm11th \: Term =a_{11} = 43$

$\large \blue \bigstar\red{\rm As\: \rm a_{11} = 43}$

$:\longmapsto \rm a + 10d = 43$

$:\longmapsto \rm 3 + 10d = 43$

$:\longmapsto \rm 10d = 43 - 3$

$:\longmapsto \rm 10d = 40$

$:\longmapsto \rm d = \cancel \frac{40}{10}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf d = 4} }}}$

Therefore,

$\Large\underline{\pink{\underline{\frak{\pmb{Common \: Difference = 4 }}}}}$

Answer 2

• We have,

11th term = 43

First term = 3

• We have to find the common difference.

We are using the formula

${ \large\sf{\red{\underline{a_n = a + (n-1)d{ \tiny \pink{ \:(nth \: term \: of \: an \: AP)}}}}}}$

★ Putting values in the formula

$\sf\rightsquigarrow \: \: {3 + 10d = 43}$

$\sf\rightsquigarrow \: \: {10d = 43 - 3}$

$\sf\rightsquigarrow \: \: 10d = 40{}$

$\sf\rightsquigarrow \: \: {d = \cancel\frac {40^4}{10_1}}$

$\sf\rightsquigarrow \: \:{d ={4}}$

${\sf{\green{\underline{\underbrace {Hence,\: the \: common \: difference \: is \: 4}}}}}$