Question

if 3 ,7,11....403 is an ap find the sum of the terms​

Answer 1

$\Huge{\underline{\underline{\blue{\mathfrak{Answer :}}}}}$

$\Large{\sf{Given :}}$

A. P :- 3, 7, 11 .......... 403

First term (a) = 3

Common Difference = 4

Last term (L or An) = 403

$\rule{200}{2}$

$\LARGE{\sf{To \: Find :}}$

Sum of all the terms

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$\LARGE{\sf{Solution :}}$

We know that,

$\LARGE{\boxed{\boxed{\green{\tt{A_{n}= a + (n - 1)d}}}}}$

(Putting Values)

403 = 3 + (n - 1)4

⇒403 - 3 = (n - 1)4

⇒400 = (n - 1)4

⇒400/4 = n - 1

⇒100 = n - 1

⇒100 + 1 = n

⇒ n = 101

$\Large{\boxed{\red{\sf{n = 101}}}}$

Now,

$\rule{200}{2}$

$\LARGE{\boxed{\boxed{\green{\tt{S_{n} = \frac{n}{2} (a + L) }}}}}$

Sn = 101 * (3 + 403) / 2

⇒Sn = 101 * 406 / 2

⇒ Sn = 101 * 203

⇒ Sn = 20503

$\Large{\boxed{\red{\sf{S_{n} = 20503}}}}$

Answer 2

Given ,

The given AP is 3 , 7 , 11 , ........ , 403

Here ,

First term (a) = 3

Common difference (d) = 7 - 3 = 4

Nth term or Last term (an or l) = 403

We know that , the nth term of an AP is given by :

$\star \: \: \sf \underline{\fbox{ \pink{a_{n} = a + (n - 1)d} }}$

Substitute the values , we obtain

$\sf\hookrightarrow403 = 3 + (n - 1)4 \\ \\ \sf \hookrightarrow 400 = (n - 1)4 \\ \\ \sf \hookrightarrow 100 = n - 1 \\ \\ \sf\hookrightarrow n = 101$

Now , The sum of the first n terms of an AP is given by :

$\fbox \green{ \: \: \star \: \: } \: \sf \: \underline{ \fbox{ \red{ \: \: S = \frac{n}{2} (a + l) \: \: }}}$

Again , substitute the values , we obtain

$\sf \hookrightarrow S = \frac{101}{2} (3 + 403) \\ \\ \sf \hookrightarrow S = \frac{101}{ \cancel{2}} ( \cancel{406}) \\ \\ \sf \hookrightarrow \sf S = 101 \times 203 \\ \\ \sf \hookrightarrow S = 20503$

Therefore , the sum of first 101 terms of given AP is 20503