Question

What is the no. of terms in the sequence 5,8,11....if its sum is 735​

Answer 1

$\huge\star{\tt{\underline{\underline{\red{Answer}}}}}\star$

AP $\implies$ 5,8,11.......n

Sum of AP = 735

a= 5

d= 8-5= 3

We dont know the Last Term

So by Formula

Last Term(a$_{n}$) = a+ (n-1)d

$\implies$ a$_{n}$=5 +( n-1)3

$\implies$ a$_{n}$= 5+3n-3 _______(1)

Sum Of Terms = $\frac{n}{2}$(a + a$_{n}$)

now from (1)

$\implies$ 735=$\frac{n}{2}$ (5 + (5 + 3n -3))

$\implies$ 735 x 2 = n(7 +3n)

$\implies$ 1470= 7n +3n²

$\implies$ 3n²+7n -1470

By Quadratic Formula

$\huge{\frac{-b±{\sqrt{{b}^{2} -4ac}}}{2a}}$

Here we have

a= 3 ,b= 7, c = -1470

so by applying it in this formula

we get

$\large{\frac{-7±{\sqrt{{7}^{2} -4 x 3 x (-1470)}}}{2 x 3}}$

$\implies$ $\large{\frac{-7±{\sqrt{49+17640}}}{6}}$

$\implies$$\large{\frac{-7±{\sqrt{17689}}}{6}}$

$\implies$$\large{\frac{-7±133}{6}}$

Two solutions we get

$\implies$$\large{\frac{-7-133}{6}}$

$\implies$ -23.33

but we know that no. of terms are positive and a natural no.

so

$\implies$$\large{\frac{-7+133}{6}}$

$\implies$ $\large{\frac{126}{6}}$

$\star$$\huge{\boxed{\boxed{21}}}$