Question

For an AP, a3 = 7, d = 4, sn = 740, find n and an


Plz solve full Q3 ​

Answer 1

Given:

• $\sf a_{3} = 7$
• $\sf d = 4$
• $\sf S_{n} = 740$

To Find:

• $\sf n \: and \: a_{n}$

Solution :

$\sf \: \sf \longrightarrow \: a_{3} = 7 \\ \sf \: \sf \longrightarrow \: a + 2d = 7 \\ \sf \longrightarrow\sf \: a + 2(4) = 7 \: ... \{ \because \: d = 4 \} \\ \sf \longrightarrow \sf \: a = 7 - 6 \\ \sf \: \sf \longrightarrow \: a = 1 \\ \sf \longrightarrow \small{\boxed{\sf \: a_{n} = a = 1 }}$

Now,

$\sf \longrightarrow \: </u></strong><strong><u>S</u></strong><strong><u>_{n} = \dfrac{n}{2} \{2a + (n - 1)d \} \\ \\ \sf \longrightarrow740 = \dfrac{n}{2} \{2(1) + (n - 1)4 \} \\ \\ \sf \longrightarrow \: 740 = \dfrac{n}{2} (2 + 4n - 4 )\\ \\ \sf \longrightarrow \: 740 = \dfrac{n}{2} (4n - 2) \\ \\ \sf \longrightarrow \: 1480 = n(4n - 2)\\ \\ \sf \longrightarrow \: \small{\boxed{ \sf \: n = \frac{1480}{4n - 2} }}$