Question

find the sum of first 15 terms of ap 5,8,11,14.......​

Answer 1

$\sf\huge\blue{\underline{\underline{ Question : }}}$

Find the sum of first 15 terms of ap 5,8,11,14.......

$\sf\huge\blue{\underline{\underline{ Solution : }}}$

★ Given that,

• AP : 5,8,11,14.....

★ To find,

• Sum of first 15 terms.

★ Let,

$\tt\:a_{1} = 5$

$\tt\:a_{2} = 8$

Common difference (d) = a2 - a1

➡ 8 - 5 = 3

$\boxed{Common\:difference = 3 }$

★ Now,

• a = 5
• d = 3
• n = 15

By using sum of nth terms of an AP :

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

Where,

• a = First Term of AP.
• d = Common Difference of AP.
• n = Number of terms of AP.
• Sn = Sum of nth terms of AP.

$\bf\:\leadsto S_{15} = \frac{15}{2} [ 2(5) + (15 - 1)(3) ]$

$\bf\:\leadsto S_{15} = \frac{15}{2} [ 10 + (14)(3) ]$

$\bf\:\leadsto S_{15} = \frac{15}{2} [ 10 + 42 ]$

$\bf\:\leadsto S_{15} = \frac{15}{2} [ 52 ]$

$\bf\:\leadsto S_{15} = 15[26]$

$\bf\:\leadsto S_{15} = 390$

$\underline{\boxed{\bf{\purple{ \therefore Sum\:of\:15\:terms\:of\:an\:AP= 390.}}}}\:\orange{\bigstar}$

★ More Information :

$\boxed{\begin{minipage}{5 cm} AP Formulae \\ \\ : \implies a_{n} = a + (n - 1)d \\ \\ :\implies S_{n} = \frac{n}{2} [ 2a + (n - 1)d ] \end{minipage}}$

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Answer 2

Answer: 390

Step-by-step explanation:

Sn=n/2[2a+(n-1)d]

n=15, a=5, d=3

S15=15/2[2*5+(15-1)3]

S15=15/2[10+42]

S15=15/2[52]

S15=15*26

S15=390 ans.