Question

The 20th term of an A.Pis 58 and the sum of 20 term is 590. Find the common difference and sum of first to terms
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Answer 1

GIVEN :–

• 20th term of A.P. = 58

• Sum of 20 term = 590

TO FIND :–

• Common difference = ?

• Sum of n terms = ?

SOLUTION :–

• We know that nth term of A.P. –

$\\ \implies\sf T_n = a + (n - 1)d$

• Now put n = 20 –

$\\ \implies\sf T_{20}= a + (20 - 1)d$

$\\ \implies\sf 58= a +19d \: \: \: \: - - - eq.(1)$

• We also know that sum of n terms –

$\\ \implies\sf S_n = \dfrac{n}{2} [2a + (n - 1)d]$

• Put n = 20 –

$\\ \implies\sf S_{20} = \dfrac{20}{2} [2a + (20 - 1)d]$

$\\ \implies\sf 590=(10)(2a +19d)$

$\\ \implies\sf \dfrac{590}{10}=2a +19d$

$\\ \implies\sf 59=2a +19d$

$\\ \implies\sf 59=a + a +19d$

• Now using eq.(1) –

$\\ \implies\sf 59=a + 58$

$\\ \implies\sf a = 59 - 58$

$\\ \implies\sf a =1$

• Put the value of 'a' in eq.(1) –

$\\ \implies\sf 58= 1+19d$

$\\ \implies\sf 58 - 1 = 19d$

$\\ \implies\sf 57 = 19d$

$\\ \implies\sf d = \cancel \dfrac{57}{19}$

$\\ \large\implies{ \boxed{\sf d =3}}$

▪︎Hence , Common difference is 3.

▪︎Now Let's find sum of n terms –

$\\ \implies\sf S_n = \dfrac{n}{2} [2a + (n - 1)d]$

$\\ \implies\sf S_n = \dfrac{n}{2} [2(1)+ (n - 1)3]$

$\\ \implies\sf S_n = \dfrac{n}{2} (2+3n - 3)$

$\\ \implies\sf S_n = \dfrac{n}{2} (3n - 1)$

$\\ \large\implies{ \boxed{\sf S_n = \dfrac{3}{2} {n}^{2} - \dfrac{n}{2}}}$

Answer 2

Given :

• The 20th term of an A.P is 58

• the sum of 20 term is 59.

To Find :

• Find the common difference and sum of first to terms

Solution :

$\underline \boldsymbol{According \: to \: the \: Formulas } \: :$

$: \implies \: \: \: \: \: \boxed{ \sf \: t_{n} = a +(n - 1)d }$

The 20th term of an A.P is 58 :

$\sf \: : \implies \: \: \: \: \: \: \: \: \: \: 58 = 1 + 19d \\ \\ \\ \sf \: : \implies \: \: \: \: \: \: \: \: \: \:19d = 58 - 1 \\ \\ \\ \sf \: : \implies \: \: \: \: \: \: \: \: \: \:19d = 57 \\ \\ \\ \sf \: : \implies \: \: \: \: \: \: \: \: \: \:d = \cancel{\frac{57}{19} } \\ \\ \\ \sf \: : \implies \: \: \: \: \: \: \: \: \: \:d = 3$

the sum of 20 term is 59 :

$\sf \: : \implies \: \: \: \: \: \: \: \: \: \: 59 = a + 19d$

Substitute all values :

$\sf \: : \implies \: \: \: \: \: \: \: \: \: \: 59 = a \: + 19 \times 3 \\ \\ \\ \sf \: : \implies \: \: \: \: \: \: \: \: \: \: 59 \: = a +58 \\ \\ \\ \sf \: : \implies \: \: \: \: \: \: \: \: \: \: a = 59 - 58 \\ \\ \\ \sf \: : \implies \: \: \: \: \: \: \: \: \: \: a = 1$

Hence the a is 1 and D is 3