Question

Find the sum of first 20 terms of the sequence 5, 10, 15,​

Answer 1

EXPLANATION.

Sequence = 5, 10, 15,,,,,,

As we know that,

Sum of Nth term of an A.P.

⇒ Sₙ = n/2[2a + (n - 1)d].

First term = a = 5.

Common difference = d = b - a = c - b.

Common difference = d = 10 - 5 = 15 - 10.

Common difference = d = 5.

⇒ S₂₀ = 20/2[2(5) + (20 - 1)(5)].

⇒ S₂₀ = 10[10 + (19)(5)].

⇒ S₂₀ = 10[10 + 95].

⇒ S₂₀ = 10[105].

⇒ S₂₀ = 1050.

                                                                                                                     

MORE INFORMATION.

Arithmetic progression.

If a is the first term and d is the common difference then A.P. can be written as : a, a + d, a + 2d, a + 3d + ,,,,,

General term of an A.P.

General term (nth term) of an A.P. is given by,

Tₙ = a + (n - 1)d.

Answer 2

GIVEN :-

• AP is 5,10,15...

TO FIND :-

• Sum of first 20 terms.

TO KNOW :-

$\\ \bigstar\boxed{ \sf \: s_{n} = \frac{n}{2} \{2a + (n - 1)d \}}$

Here ,

• s(n) = Sum of 'n' terms.
• n = Number of terms.
• a = 1st term
• d = Common difference

SOLUTION :-

We will find Common difference (d) ,

→ d = 2nd term - 1st term

→ d = 10 - 5

→ d = 5

____________________

♦ First term {a} = 5

♦ Common difference {d} = 5

♦ Number of terms {n} = 20

Putting values in formula ,

$\\ \sf \: s_{n} = \frac{n}{2} \{2a + (n - 1)d \} \\ \\ \sf s_{20} = \cancel{\frac {20}{2}} \{2(5) + (20 - 1)5 \} \\ \\ \: \: \: \: \: \implies \sf 10 \{10 + (19)5 \} \\ \\ \: \: \: \: \: \implies \sf 10(10 + 95) \\ \\ \: \: \: \: \: \implies \sf 10(105) \\ \\ \: \: \: \: \: \implies \boxed{ \bf \: s_{ 20} =1050 }$

Hence , sum of first 20 terms is 1050.

MORE TO KNOW :-

'n'th term of AP is given as,

a(n) = a + (n - 1)d

Here ,

• a(n) = "n"th term
• a = 1st term
• n = Total terms
• d = Common difference