Question

In an AP S5=35 and S1=22 find 5th term

Answer 1

ശ്ജ്ബക്വ്വ്കക്യൂവജ്ബക്ന്ന

Answer 2

Step-by-step explanation:

${S}^{5} = 35$

$= > \frac{n}{2} [2a + (n - 1)d] = 35$

$= > \frac{5}{2} [2a + 4d] = 35$

$= > 2a + 4d = 35 \times \frac{2}{5}$

$= > 2a + 4d = 7 \times 2$

$= > 2a + 4d = 14 \: \: \: \: \: \: \: \: \: \: --------1$

${S}^{4} = 22$

$= > \frac{4}{2} [2a + (4 - 1)d] = 22$

$= > 2[2a + 3d] = 22$

$= > 2a + 3d = 11 \: \: \: \: \: \: \: \: \: \: -------- \: 2$

On subtracting equation 1 and 2, we get

d = 3

Now

On substituting the value of d in equation 1, we get

$= > 2a + 4 \times 3 = 14$

$= > 2a + 12 = 14$

$= > 2a = 2$

$= > a = 1$

$5th\:term = a + 4d$

$= 3 + 4 \times 2$

$= 3 + 8$

$= 11$