Let S50-1-23-4-56-7-89- upto 50 terms- then find S5025-1980-

Given S _ 50 =1+(2)(3)+4+(5)(6)+... upto #160; 50 #160; #160;terms ⇒ S _ 50 =(1+4+7+10+.... upto #160; #160;25 #160; terms) +(2.3 ...

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Byju's Answer

Standard X

Mathematics

Collinearity Condition

Let S50=1+2...

Question

Let $S_{50} = 1 + (2) (3) + 4 + (5) (6) + 7 + (8) (9) + . . .$ upto $50$ terms, then find $\frac{S_{50}}{25} - 1980$ .

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Solution

Given $S_{50} = 1 + (2) (3) + 4 + (5) (6) + . . .$ upto 50 terms
$\Rightarrow S_{50} = (1 + 4 + 7 + 10 + . . . .$ upto 25 terms) $+ (2.3 + 5.6 + 8.9 + . . . .$ 25 terms)
or $S_{50} = \sum_{r = 1}^{25} (3 r - 2) + \sum_{r = 1}^{25} 3 r (3 r - 1)$
${\Rightarrow S}_{50} = 49675$
Now, $\frac{S_{50}}{25} - 1980 = \frac{49675}{25} - 1980$
$\Rightarrow \frac{S_{50}}{25} - 1980 = \frac{49675 - 49500}{25}$
$\Rightarrow \frac{S_{50}}{25} - 1980 = \frac{175}{25}$
$\Rightarrow \frac{S_{50}}{25} - 1980 = 7$

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Let S50=1+(2)(3)+4+(5)(6)+7+(8)(9)+... upto 50 terms, then find S5025−1980.

Given S50=1+(2)(3)+4+(5)(6)+...upto 50 terms⇒S50=(1+4+7+10+....upto 25 terms)+(2.3+5.6+8.9+....25 terms)or S50=∑25r=1(3r−2)+∑25r=13r(3r−1)⇒S50=49675Now, S5025−1980=4967525−1980⇒S5025−1980=49675−4950025⇒S5025−1980=17525⇒S5025−1980=7

Let $S_{50} = 1 + (2) (3) + 4 + (5) (6) + 7 + (8) (9) + . . .$ upto $50$ terms, then find $\frac{S_{50}}{25} - 1980$ .