Question

If fore on arithmetic progression the first term is 3 and the common difference is 6 then find S10

Answer 1

GiveN :

• First term (a) = 3
• Common Difference (d) = 6

To FinD :

• Sum of 10 terms

SolutioN :

Use formula of sum of terms

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

ㅤㅤㅤㅤㅤㅤㅤ

Put n = 10

ㅤㅤㅤㅤㅤㅤㅤ

$\implies$ S10 = 10/2[2(3) + (10 - 1)6]

ㅤㅤㅤㅤㅤㅤㅤ

$\implies$ S10 = 5[6 + (9)6]

ㅤㅤㅤㅤㅤㅤㅤ

$\implies$ S10 = 5[6 + 54]

ㅤㅤㅤㅤㅤㅤㅤ

$\implies$ S10 = 5[60]

ㅤㅤㅤㅤㅤㅤㅤ

$\implies$ S10 = 300

ㅤㅤㅤㅤㅤㅤㅤ

$\underline{\rm{\therefore \: Sum \: of \: 10 \: terms \: of \: AP \: is \: 300}}$

Answer 2

$\sf{\underline{\boxed{\green{\large{\bold{ Given}}}}}}$

• first term = a = 3
• common difference = d = 6

$\sf{\underline{\boxed{\green{\large{\bold{ To\: find}}}}}}$

• sum of 1st 10 terms = n = ??

$\sf{\underline{\boxed{\green{\large{\bold{ Solution}}}}}}$

$\sf{\underline{\boxed{\blue{\large{\bold{ s_n = \dfrac{n}{2} ( 2a + ( n - 1)d ) }}}}}}$

$\sf\implies s_{10} = \dfrac{10}{2}( 2\times 3 + ( 10 - 1 ) 6 )$

$\sf\implies s_{10} = 5 ( 6 + 9 \times 6 )$

$\sf\implies s_{10} = 5 ( 6 + 54 )$

$\sf\implies s_{10} = 5 \times 60$

$\sf\implies s_{10} = 300$

$\sf{\underline{\boxed{\pink{\large{\mathfrak{ sum \: of \: first \: 10\: terms \: = \: 300}}}}}}$