Question

the sum of first 7 terms of an arthematic sequence is 133 and the 7th term is 31 so what is the 4th term​

Answer 1

Answer:

19

Step-by-step explanation:

Sum of 1st n term = (n/2)[a + l], where a is first term and l is last term.

 Here, let the first term be 'a'.

Sum =  133

⇒ (7/2)[ a + 31] = 133  

⇒ (3.5)(a + 31) = 133

⇒ a = (133/3.5) - 31

⇒ a = 38 - 31   = 7

      Hence,

7th term = 31

a + 6d = 31    ⇒ 7 + 6d = 31

⇒ 6d = 31 - 7 = 24

⇒ d = 4

            Therefore,

4th term = a + 3d

              = 7 + 3(4)

              = 19

Answer 2

Given :-

• Sum of first 7 terms = 133

• 7th term = 31

• Number of terms = 7

To Find :-

• 4th term of the sequence

Solution :-

By using A.P formula

$\implies \sf A_n = \frac{n}{2} \big( \: a + l \: \big) \\ \\ \implies \sf133 = \frac{7}{2}\big(a + 31\big) \\ \\ \implies \sf \frac{133 \times 2}{7} = a + 31 \\ \\\implies \sf 38 = a + 31 \\ \\\implies \sf a = 38 - 31 \\ \\ \implies\underline{ \boxed{\sf a =7}}$

Now

$\implies \sf A_7 = 31 \\ \\ \implies \sf a + 6d = 31 \\ \\ \implies \sf 7 + 6d = 31 \\ \\ \implies \sf 6d = 31 - 7 \\ \\ \implies \sf 6d = 24 \\ \\ \implies \sf d = \frac{24}{6} \\ \\ \implies\underline{\boxed{ \sf d = 4}}$

Now 4th term of the sequence

$\implies \sf A_4 = a + 3d \\ \\ \implies \sf A_4 = 7 + 3 \times 4 \\ \\ \implies \sf A_4 = 7 + 12 \\ \\ \implies \underline{ \boxed{ \sf A_4 = 19}}$