Question

The number of the Terms of series
10+29/3 +28/3+9+. ...will amount
to 155 is ... ?​

Answer 1

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FORMULA TO BE IMPLEMENTED

Sum of first n terms of an arithmetic progression

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

Where First term = a

Common Difference = d

TO DETERMINE

The number of the Terms of series

$\displaystyle \: 10 + \frac{29}{3} + \frac{28}{3} + 9 + ...........$

will amount to 155

EVALUATION

Let n be the number of terms

This is an Arithmetic progression

$First \: \: term = 10$

$\displaystyle \: Common \: \: Difference = \frac{29}{3} - 10 = \frac{29 - 30}{3} = - \frac{1}{3}$

So by the given condition

$\displaystyle \: \frac{n}{2} [(2 \times 10) + (n - 1) \times \frac{ - 1}{3} ] = 155$

$\implies \: \: \displaystyle \: \frac{n}{2} [ \: 20 - \frac{ (n - 1) }{3} ] = 155$

$\implies \: \: \displaystyle \: n(61 - n) = 930$

$\implies \: \: \displaystyle \: {n }^{2} - 61n + 930 = 0$

$\implies \: \: \displaystyle \: {n }^{2} - 30n - 31n + 930 = 0$

$\implies \: \: \displaystyle \: (n - 30)(n - 31) = 0$

So

$n = 30 \: \: , \: \: 31$

RESULT

The required number of terms = 30 and 31

EXPLANATION OF TWO ANSWERS

Since the 31 th term of the above Arithmetic Progression is

$= \displaystyle \: a + (n - 1)d$

$= \displaystyle \: 10 + \bigg[ (31 - 1) \times ( - \frac{1}{3} ) \bigg]$

$= 10 - 10$

$= 0$

For this reason there is two answers