Question

If the Sum of the first 3 terms of an AP is 10 and that of the first four terms is 18 then the fourth term of the AP is?​

Answer 1

Answer:

8

Step-by-step explanation:

a1 + a2 + a3 = 10

a1 + a2 + a3 + a4 = 18

So a4 = 18 - 10 = 8

Answer 2

       $\underline{\bold{Solution:}}$

       $\text{It is given that the sum of the first 3 terms of an AP is 10}$

       $\text{We know that, sum of }n \text{ terms of an AP is}(S_n) = \dfrac{n}{2}[2a+(n-1)d]$

       $\text{Where }a \text{ is the first term of AP and }d \text{ is the common difference}$

$\implies S_3 = \dfrac32[2a + (3-1)d] = 10$$2(1) + 3d = 9$

$\implies S_3 = \dfrac32(2a + 2d) = 10$

$\implies S_3 = 3(a+d) = 10$

$\implies 3a + 3d = 10 \text{ ------ (i)}$

       $\text{It is also given that the sum of the first 4 terms of the AP is 18}$

$\implies S_4 = \dfrac42[2a + (4-1)d] = 18$

$\implies S_4 = 2[2a + 3d] = 18$

$\implies2a + 3d = 9 \text{ ------ (ii)}$

       $\text{Substracting (ii) from (i)}$

$\implies 3a + 3d - 2a - 3d = 10 - 9$

$\implies a = 1$

       $\text{Substituting in (ii)}$

$\implies 2(1) + 3d = 9$

$\implies 3d = 9 -2$

$\implies d = \dfrac{7}{3}$

       $\text{We know that }n^{th} \text{ term of an AP}(a_n) = a + (n-1)d$

$\implies a_4 = 1 + (4-1)(\dfrac73)$

$\implies a_4 = 1 + 3(\dfrac73)$

$\implies a_4 = 1 + 7$

$\implies \boxed{a_4 = 8}$