Question

How many terms
8 +11 +14 ... +
of series
will may total
1660​

Answer 1

Answer:

32 , 34

Step-by-step explanation:

Since the given series is;

8+11+14+........+

The total sum of the series is  1660

Since the terms are in Arithmatic Progression.

Here the first term is (a) = 8

The common difference (d) = 11-8=3=14-11

So, the formula for total sum is:

Sn = $\frac{n}{2} \times [a+(n-1)\times d]$

$1660 = \frac{n}{2} \times [ 8 + (n-1) \times 3]\\3220 = n \times [8 +  3n - 3]\\3220= n \times [5 + 3n]\\3220 = 5 \times n + 3 \times n^{2} \\3n^{2} +5n - 3220 = 0$

Applying Sreedharacharya method to find n;

$d = b^{2} - 4 \times a \times c\\d= 5^{2} - 4 \times 3 \times (-3220)\\d= 25 + 38640\\d = 38665$

Now finding n;

$n = \frac{ -b + \sqrt{d} }{2 \times a}$

$n = \frac{-5 + \sqrt{38665} }{6}$

$n = \frac{196.6-5}{6}$

n = 31.9

∴ n = 32

again,  $n = \frac{ -b - \sqrt{d} }{2 \times a}$

$n = \frac{-5-\sqrt{38665} }{6}$

$n = \frac{ -5-196.6}{6}$$n = \frac{201.6}{6}$

n= 33.6

∴ n = 34