Question

In an A.p the first term is - 5 and last
term is 45. If Sum of all numbers in the
Ap is 120, then now many terms are
there? what is the Common difference? ?​

Answer 1

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

$\:\:$

$\large\underline{\green{\bf Given :-}}$

$\:\:$

• The first term is - 5

$\:\:$

• Last term is 45

$\:\:$

• Sum of all numbers in the AP is 120

$\:\:$

$\large\underline{\red{\bf To \: Find :-}}$

$\:\:$

• Number of terms

$\:\:$

• Common difference

$\:\:$

$\large\underline{\orange{\bf Solution :-}}$

$\:\:$

We know that ,

$\:\:$

➠ $\sf a_n = a + (n - 1)d$ ⚊⚊⚊⚊ ⓵

$\:\:$

Where,

$\:\:$

• $\sf a_n$ = nth term

• a = First term

• n = Number of term

• d = Common difference

$\:\:$

$\underline{\bold{\texttt{For last term :}}}$

$\:\:$

• $\sf a_n$ = 45

• a = -5

• n = n

• d = d

$\:\:$

⟮ Putting these values in ⓵ ⟯

$\:\:$

➜ $\sf 45 = -5 + (n - 1)d$

$\:\:$

➜ 45 + 5 = (n - 1)d

$\:\:$

➜ 50 = (n - 1)d ⚊⚊⚊⚊ ⓶

$\:\:$

$\underline{\bold{\texttt{Sum of terms in AP :}}}$

$\:\:$

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

$\:\:$

Or,

$\:\:$

➠ $\sf S_n = \dfrac { n } { 2 } (a + l)$ ⚊⚊⚊⚊ ⓷

$\:\:$

Where,

$\:\:$

• $\sf S_n$ = Sum of n terms

• n = Number of terms

• a = First term

• d = Common difference

• l = Last term

$\:\:$

$\underline{\bold{\texttt{For sum of all numbers in the given AP :}}}$

$\:\:$

• $\sf S_n$ = 120

• a = -5

• l = 45

$\:\:$

⟮ Putting these values in equation ⓷ ⟯

$\:\:$

➜ $\sf S_n = \dfrac { n } { 2 } (a + l)$

$\:\:$

➜ $\sf 120 = \dfrac { n } { 2 } (-5 + 45)$

$\:\:$

➜ 120 × 2 = n(40)

$\:\:$

➜ 240 = 40n

$\:\:$

➜ $\sf n = \dfrac { 240 } { 40 }$

$\:\:$

➨ n = 6 ⚊⚊⚊⚊ ⓸

$\:\:$

• Hence there are 6 terms in the given AP

$\:\:$

⟮ Putting n = 6 from ⓸ to ⓶ ⟯

$\:\:$

➜ 50 = (n - 1)d

$\:\:$

➜ 50 = (6 - 1)d

$\:\:$

➜ 50 = 5d

$\:\:$

➜ $\sf d = \dfrac { 50 } { 5 }$

$\:\:$

➨ d = 10

$\:\:$

• Hence the Common difference is 10

$\:\:$

∴ The number of terms and the common difference is 6 & 10 respectively

$\:\:$

═════════════════════════

Answer 2

$\huge\underline{\bf \orange{AnSweR :}}$

We know that ,

➠ \sf a_n = a + (n - 1)da n =a+(n−1)d ⚊⚊⚊⚊ ⓵

Where,

n = nth term

a = First term

n = Number of term

d = Common difference

For last term :

n = 45

a = -5

n = n

d = d

⟮ Putting these values in ⓵ ⟯

➜ \sf 45 = -5 + (n - 1)d45=−5+(n−1)d

➜ 45 + 5 = (n - 1)d

➜ 50 = (n - 1)d ⚊⚊⚊⚊ ⓶

Sum of terms in AP :

➠ = 2n (2a+(n−1)d)

Or,

➠ n = 2n (a+l) ⚊⚊⚊⚊ ⓷

Where,

n = Sum of n terms

n = Number of terms

a = First term

d = Common difference

l = Last term

For sum of all numbers in the given AP :

n = 120

a = -5

l = 45

⟮ Putting these values in equation ⓷ ⟯

➜ \sf S_n = \dfrac { n } { 2 } (a + l)S n = 2n(a+l)

➜ \sf 120 = \dfrac { n } { 2 } (-5 + 45)120= 2n (−5+45)

➜ 120 × 2 = n(40)

➜ 240 = 40n

➜ \sf n = \dfrac { 240 } { 40 }n= 40240

➨ n = 6 ⚊⚊⚊⚊ ⓸

Hence there are 6 terms in the given AP

⟮ Putting n = 6 from ⓸ to ⓶ ⟯

➜ 50 = (n - 1)d

➜ 50 = (6 - 1)d

➜ 50 = 5d

➜ 5/50d= 550

➨ d = 10

Hence the Common difference is 10

∴ The number of terms and the common difference is 6 & 10 respectively.