Question

(51 + 54 + 57 +.......+150) whole square​

Answer 1

$\huge\mathfrak\blue{Answer:}$

$\huge\orange{1,16,75,889}$

Step-by-step explanation:

To Find:

${(51 + 54 + 57 + ..... + 150)}^{2}$

Solution:

We can see that 51 , 54 , 57 ,....., 150 are in AP , with:

• $a = 51$
• $d = 54 - 51 = 3$
• $an = l = 150$
• $n = ?$

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So,

$> > an = a + (n - 1)d$

$> > 150 = 51 + (n - 1)3$

$> > 150 - 51 = (n - 1)3$

$> > 99 = (n - 1)3$

$> > n - 1 = \frac{99}{3}$

$> > n - 1 = 33$

$> > n = 33 + 1$

$> > n = 34$

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According to the Q

$>>{(51 + 54 + 57 + ..... + 150)}^{2}$

$> >( Sn) {}^{2}$

$> > [ \frac{n}{2} (a + l)] {}^{2}$

$> > [ \frac{34}{2} (51 + 150)] {}^{2}$

$> > (17 \times 201) {}^{2}$

$> > (3417) {}^{2}$

$\orange {> > 1,16,75,889}$

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