Question

find the sum of first 15 terms of an arithmetic sequence of first term 5 and common difference 4?​ sequence is 5,9,13,17 etc​

Answer 1

Step-by-step explanation:

a = t1 = 5 , d= 4 , n= 15

Sn = n [2a+(n-1)× d ]

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2

= 15 [ 2× 5+ (15-1)×4]

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2

= 15 [66]

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2

= 15 × 33

=. 495. is the answer

Answer 2

Answer:

Concept

~here the concept of Arithmetic progression is used. the first term is given as 5 and the common difference as 4. we have to find the sum of first 15 term using the formula

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Formula used

$\bigstar \bold \orange{S_n= \frac{n}{2}[2a+(n-1)×d]}$

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Solution

• First term = 5
• common difference= 4
• n= 15 terms
• Arithmetic progression= 5,9,13,17,....

Let's proceed

Subsitute the value in the formula

$\sf{\implies S_{15} =\frac{15}{2}[2×5+(15-1)×4]}\\ \\ \sf{\implies S_{15} =\frac{15}{2}[10+14×4]}\\ \\ \sf{ \implies S_{15}=\frac{15}{2}×66}\\ \\ \sf {\implies\color{purple}S_{15}=495}$

therefore the sum of first 15 terms of given Arithmetic progression is 495.

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More to know !!

$\sf{ \bull\: Sum \: of \: the \: terms \: if \: last \: term \: is \: given}$

$\sf{ \to S_n = \frac{n}{2}[ a+L]}$

$\sf{ \bull \: General \: Form \: of \: Ap}$

$\sf{ \to \: a, a + d, a + 2d, a + 3d, . . }$

$\sf{ \bull \: The \: nth \: term \: of \: AP }$

$\sf{ \to \: a_ n= a + (n – 1) × d}$

where

• a denotes first term
• d denotes common difference
• n denotes the terms
• L denotes the last term

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