Question

The number of the terms of the series 10+9 2/3+9 1/3+9+........will amount to 155 is?

Answer 1

155 = n/2 ( 2× 10 - ( n-1) 1/3)
310= n ( 20 - ( n-1) 1/3)

930 = n( 60 - n +1)
930 = 60 n - n.n + n

930 = 61 n - n.n

n.n - 61n +930 = 0

n = (61 +- 1)/2
= 31,30

Answer 2

At 30 and 31 terms the sum of the terms will be amount to 155

Given:

10+9 2/3+9 1/3+9+........is a number series

To find:

The number of terms at which 10+9 2/3+9 1/3+9+........will amount to 155

Solution:

Given sequence 10+9 2/3+9 1/3+9+...... is a Arithmetic sequence

Where first term a = 10

and common difference d =T₂ - T₁ = 9 2/3 - 10 = -1/3

⇒ Common difference d = -1/3  

Let's assume that at n terms the sum of the terms = 155

As we know Sum of n terms = $\frac{n}{2} [ 2a+(n-1)d ]$  

⇒ $\frac{n}{2} [ 2a+(n-1)d ] = 155$

⇒ $n [ 2a+(n-1)d ] = 155 (2)$

⇒ $n [ 2a+(n-1)d ] = 310$

From above data, a = 10  and d = -1/3

⇒ $n [ 2(10)+(n-1)(-\frac{1}{3} ) ] = 310$

⇒ $n [ 20-\frac{n}{3} +\frac{1}{3} ] = 310$

⇒  $n [ \frac{60 -n+1}{3} ] = 310$

⇒ $n [ 60 -n+1] = 930$

⇒ 60n - n² + n = 930

⇒ n² - 61n + 930 = 0

Now fectorize  n²- 61n + 930 = 0 to get n value

⇒ n²- 61n + 930 = 0

Split -61n as -30n - 31n

⇒ n²- 30n - 31n + 930 = 0

⇒ n(n-30) -31 (n-30) = 0

⇒  (n-30)(n-31) = 0

⇒  n - 30 = 0     and      n - 31 = 0

⇒  n = 30           and      n = 31

Therefore,

At 30 and 31 terms the sum of the terms will be amount to 155

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