Question

the eleventh term of an arithmetic sequence is 30 and the sum of the first eleven terms is 55. what is the common difference?​

Answer 1

The value of Common Difference(d) = 5.

Step-by-step explanation:

★ Given that :

◼ Eleventh term of an AP series : 30

• a11 = 30

◼ Sum of first eleven terms of an AP series ; 55

• S11 = 55

★ To find :

◼ The common difference (d).

★ Let :

$\sf \rightarrow a_{11} : a + 10d = 30\:.....(1)$

$\sf \rightarrow S_{n} = \cfrac{n}{2} \bigg [ 2a + (n-1) d\bigg ]$

• Sn = 55
• n = 11

$\sf \implies 55 = \cfrac{11}{2} \bigg [ 2a + (11-1) d\bigg ]$

$\sf \implies \cancel{55} \times \cfrac{2}{\cancel{11}} = 2a + 10d$

$\sf \implies 2a + 10d = 10$

$\sf \implies 2(a + 5d) = 10$

$\sf \implies a + 5d= \cancel{\cfrac{10}{2}}$

$\sf \implies a + 5d=5 \:.....(2)$

• Subtract equations (1) & (2).

$\sf \implies 5d = 25$

$\sf \implies d = \cancel{\cfrac{25}{5}}$

$\sf \implies d = 5$

• Put the value of d in (1) to get value of a.

$\sf \implies a + 10(5)=30$

$\sf \implies a + 50=30$

$\sf \implies a=30-50$

$\sf \implies a=-20$

★ Verification :

Substitute the values of a & d in (1) to get LHS = RHS.

$\sf \implies -20+10(5)$

$\sf \implies -20+50$

$\sf \implies -30$

☯ Since, LHS = RHS.

☯ Hence, it was verified.

$\underline{\boxed{\rm{\purple{\therefore Value\:of\:Common\:difference(d)\:=5.}}}}\:\orange{\bigstar}$

★ More info :

Formulae related to Arithmetic Progression :-

$\tt\red{\underline{\underline{\blacksquare \: a_{n} = a + (n -1)d}}}$

$\tt\red{\underline{\underline{\blacksquare \: S_{n} = \cfrac{n}{2} \bigg[a + (n -1)d \bigg]}}}$

Where,

• a = First term of AP.
• d = Common difference of AP.
• n = number of terms in AP.
• an = Last term of AP.
• Sn = Sum of nth terms of AP.

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