Question

The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms
and the common difference.
The first and the last terms of an AP are 17 and 350 reserti​

Answer 1

Correct question :-

The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Given :-

• $\rm{First\:term\:of\:AP\:is\:5}$

• $\rm{Last\:term\:of\:AP\:is\:45}$

• $\rm{Sum\:of\:AP\:is\:400}$

To Find :-

• $\rm{Number\:of\:terms\:in\:AP}$

• $\rm{And\:there\:common\: difference}$

Solution :-

$\large{\boxed{\rm{\implies\: Last \: term = a_1 + (n-1) d }}}$

$\large{\boxed{\rm{\implies \: S_n = \frac{n}{2}[ a + l ] }}}$

Using 2nd Formula :-

$\rm{\implies 400 = \frac{n}{2} [ 5 + 45] }$

$\rm{\implies 400 = \frac{50 n}{2 } }$

$\rm{\implies 400 = \frac{ \cancel{50n}}{ \cancel{2}} \: = \: \frac{25n}{1} }$

$\rm{\implies n = \frac{400}{25} = 16}$

$\underline{\underline{\rm{Number\: of \: terms \: is \: 16}}}$

Now using 1st formula :-

$\rm{\implies 45 = 5 + (16-1) d }$

$\rm{\implies 45 = 5 + 15d }$

$\rm{\implies 45 - 5 = 15d }$

$\rm{\implies 40 = 15d }$

$\rm{\implies d = \frac{40}{15} = \frac{8}{3} }$

$\underline{\underline{\rm{Common\: difference\: \: is \: \frac{8}{3}}}}$

Answer 2

Step-by-step explanation:

hi

hope this helps you

Sn=n/2(a+l)

400=n/2(5+45)

400=n/2(50)

8=n/2

therefore n=16

now,

An=a+(n-1)d

45=5+(16-1)d

40=15d

d=8/3

therefore number of terms is 16 and common difference is 8/3