Question

the sum of first seven terms of an AP is 63 and the sum of next seven is 161. find the 20th term of this AP

Answer 1

Answer:41

Step-by-step explanation:

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Answer 2

We know that, Sum of 'n' terms of an A.P is given by :

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

Where :

✡  'a' is the First term of the A.P

✡  'd' is the Common difference of the A.P

Given : Sum of First Seven terms of the A.P is 63

$\bf{\implies S_7 = \frac{7}{2}[2a + (7 - 1)d]}$

$\bf{\implies \frac{7}{2}[2a + 6d] = 63}$

$\bf{\implies \frac{1}{2}(2)[a + 3d] = \frac{63}{7}}$

$\bf{\implies a + 3d = 9}$ ---------------- [1]

Given : The Sum of next Seven terms is 161

We need to realize that : In the Sum of next Seven terms, The First term will be the 8th term of the Given A.P

We know that, nth term of an A.P is given by :

✿  $\bf{T_n = a + (n - 1)d}$

$\bf{\implies 8^t^h term = T_8 = a + (8 - 1)d = [a + 7d]}$

⇒ First term of the Series of the Next 7 terms is : (a + 7d)

$\bf{\implies S_n_e_x_t_(_7_) = \frac{7}{2}[2(a + 7d) + (7 - 1)d]}$

$\bf{\implies \frac{7}{2}[2(a + 7d) + 6d] = 161}$

$\bf{\implies \frac{1}{2}(2)[a + 7d + 3d] = \frac{161}{7}}$

$\bf{\implies a + 10d = 23}$ -------------- [2]

Subtracting Equation [1] from Equation [2], We get :

⇒ (a + 10d) - (a + 3d) = 23 - 9

⇒ a + 10d - a - 3d = 14

⇒ 7d = 14

⇒ d = 2

Substituting d = 2 in Equation [1], We get :

⇒ a + 3(2) = 9

⇒ a = 9 - 6

⇒ a = 3

$\bf{\implies T_2_0 = 3 + (20 - 1)2}$

$\bf{\implies T_2_0 = 3 + (19)2}$

$\bf{\implies T_2_0 = 41}$

The 20th term of the Given A.P is 41