Question

in an AP a=15,d=3 and an=60 find n and sn​

Answer 1

Answer:

n = 16 and Sn = 600

Step-by-step explanation:

In AP,

an = a+(n-1)d

60 = 15+(n-1)3

60-15 = 3n-3

45+3 = 3n

48 = 3n

3n = 48

n = 48/3 = 16

Sn = n/2 × (a+an)

Sn = 16/2 × (15+60)

Sn = 8×75

Sn = 600

Answer 2

Given:

• In an AP,
• a [first term] = 3
• d [common difference] = 3
• a_n [last term] = 60

To find:

• n [the number of terms] and S_n [sum of the terms.]

Answer:

• Let's first find n.

$\sf The\ general\ form\ of\ a\ term\ is\ \bf a_n\ =\ a\ +\ (n\ -\ 1)d.$

Substituting the values,

$\sf 60\ =\ 3\ +\ (n\ -\ 1)3\\\\\\60\ =\ 3\ +\ 3n\ -\ 3\\\\\\60\ =\ 3n\\\\\\\dfrac{60}{3}\ =\ n\\\\\\20\ =\ n$

Therefore, the number of terms is 20.

Now, let's find the sum of the AP.

$\bf S_n\ =\ \dfrac{n}{2}\Bigg[2a\ +\ (n\ -\ 1)d\Bigg]$

Substituting the values,

$\sf S_n\ =\ \dfrac{20}{2}\Bigg[2\ \times\ 3\ +\ (20\ -\ 1)3\Bigg]\\\\\\S_n\ =\ 10\Bigg[6\ +\ 19\ \times 3\Bigg]\\\\\\S_n\ =\ 10\Bigg[6\ +\ 57\Bigg]\\\\\\S_n\ =\ 10\ \times\ 63\\\\\\S_n\ =\ 630$

Therefore, the AP has 20 terms and its sum is 630.