Question

a = 1000, d = 100, Sum of terms = 118000. find the number of terms​

Answer 1

Answer:  The number of terms " n " are 40

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Step-by-step explanation:

a = 1000 = first term , d = 100 = difference of consecutive terms

$S_{n} =118000$ = sum of " n " number of terms.

Let the number of terms be " n "

as we know that ,  

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

=>       118000 * 2 = n[ 2*1000 + ( n - 1 )100 ]

=>       236000 = 100[20n + n( n - 1 ) ]

=>       2360 = 20n + $n^{2}$ - n

=>       $n^{2} + 19n-2360=0$

=>       $n = \frac{-9+\sqrt{19^{2}-4.1.(-2360) } }{2.1}$     AND    $n = \frac{-9-\sqrt{19^{2}-4.1.(-2360) } }{2.1}$

=>       $n=\frac{-19+99}{2}$  

BUT  $n=\frac{-19-99}{2}$     , neglect as number of terms can never be negative .

.:     n = 40