Question

if t5=9 and t11=21 find the first term and the common difference.​

Answer 1

Step-by-step explanation:

t5=t1+(n-1)d

t11=t1+(n-1)d

solving it by elimination

subtracting t5 from t11

21-9=10d-8d

d=6

putting value of d in any of the equation to get t1 or first term

9=t1+8(6)

t1= -41

Answer 2

Answer:

• First Term (a or $t_{1st}$) = 1

• Common Difference (d) = 2

Given:

• $t_{5th}$ = 9

• $t_{11th}$ = 21

To Find:

• First Term and Common Difference of the AP

Solution:

We know that,

General Term of an AP = a + ( n - 1 )d

where,

a = first term of the AP

n = number of terms or the position of a term in AP

d = common difference between two terms in AP

In the question,

$t_5$ = a + ( 5 - 1 )d

9 = a + 4d .......(1)

$t_{11}$ = a + ( 11 - 1 )d

21 = a + 10d ......(2)

Subtracting equation (1) from equation (2), we get:

a + 10d - a - 4d = 21 - 9

6d = 12

d = $\dfrac{12}{6}$

d = 2

Substituting the value of common difference in equation (1), we get:

a + 4d = 9

a + 4*2 = 9

a + 8 = 9

a = 1

Therefore, the answer is:

a = 1

d = 2

✨Other AP Formulas:✨

⭐nth term of an AP

nth term of an APformulas⭐

$\\\\\sf 1) \: n_{th} \: term \: of \: any \: AP \: = a + (n - 1)d$

$\sf 2) \:n_{th} \: term \: from \: the \: end \: of \: an \: AP \: = a + (m - n)d$

$\sf 3) \:n_{th} \: term \: from \: the \: end \: of \: an \: AP = l - (n - 1)d$

$\sf 4) \: Difference \: of \: two \: terms = (m - n)d$

where m and n is the position of the term in the AP

$\\\sf 5) \:Middle\: term\: of\: a\: finite\: AP$

$\sf (i) \:\: If \: n \: is \: odd = \frac{n + 1}{2}\:th\:term$

$\sf (ii) \:\: If \: n \: is \: even = \frac{n}{2} \:th \: term \: and \: ( \frac{n}{2} + 1)th \: term$

⭐Sum Formulas⭐

$\sf 1) \: Sum \: of \: first \: n \: terms \: of \: an \:AP = \frac{n}{2} [ \: 2a + (n - 1)d \: ]$

$\sf 2) \: Sum \: of \: first \: n \: natural \: numbers = \frac{n(n + 1)}{2}$

$\sf 3) \: Sum \: of \: AP \: having \: last \: term = \frac{n}{2} [ \: a + l \: ]$