Question

find the 19th term of the following A.P , 7,13,19,25....l​

Answer 1

Given Sequence

7,13,19,25 ,,,,,,

⇒First term (a) = 7

⇒Common difference(d) = a₂-a₁= 13-7 = 6

⇒Number of term (n) = 19

To find

⇒19th term of ap

Formula

⇒Tₙ = a + (n-1)d

Now we can write as

⇒T₁₉ = 7 + (19-1)×6

⇒T₁₉ = 7 + 18×6

⇒T₁₉ = 7 + 108

⇒T₁₉ = 115

Answer

⇒T₁₉ = 115

                                                     

More Formula about Sequence and series

⇒a,b,c are in AP ; 2b = a+c , GP ; b²=ac and HP ; b = (2ac)/(a+c)

⇒AM = (a+b)/2 , G= √(ab) and H = (2ab)/(a+b)

⇒A≥G≥H        In Equality Based

Answer 2

Answer:

$\large \underline{\sf\pmb{Given}}$

$: \implies \sf{A.P = 7,13,19,25...}$

$\large \underline{\sf \pmb{To \: Find }}$

$: \implies \sf{19th \: term \: of A.P}$

$\large \underline{\sf\pmb{Using \: Formula }}$

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

Where

• ⇒ tn = nth term
• ⇒ n = Number of Term
• ⇒ a = First term
• ⇒ d = Common deference

$\large \underline {\sf \pmb{Solution}}$

$\bigstar \: \frak{Here \: we \: know \: that:}$

• ⇒ First Term = 7
• ⇒ Number of Term = 19

$\bigstar \: \frak{\: Finding \: Common \: Differences \: between \: terms}$

$: \implies\sf{a_2-a_1}$

$: \implies \sf{13 - 7}$

$: \implies \sf{6}$

• ⇒ The common difference between terms is 6.

$\bigstar \: \frak{Now \: Finding \: 19th \: Term}$

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

• ⇒ Substituting the values

$: \implies { \sf{T_{19}=7+(19-1)6}}$

$: \implies { \sf{T_{19}=7+(18 \times 6)}}$

$: \implies { \sf{T_{19}=7+108}}$

$: \implies { \sf{T_{19}=115}}$

• ⇒ Hence, The 19th term is 115.

$\large \underline{ \sf \pmb{More \: Useful \: Formulae}}$

★ Formula to find the numbers of term of an AP:

• $: \implies \sf{n= \bigg[ \dfrac{(l - a)}{d} \bigg]}$

★ Formula to find the tsum of first n terms of an AP:

• $: \implies\sf{S_n= \dfrac{n}{2} (a + l)}$

★ Formula to find the sum of squares of first n natural numbers of an AP:

• $: \implies \sf{S = \dfrac{n(n + 1)(2n + 1)}{6} }$

★ Formula to find the nth term of an AP is the square of the number of terms:

• $: \implies \sf{S = {n}^{2} }$

★ Formula to find the sum of of an AP:

• $: \implies \sf{S = n(n+1)}$