the eleventh term of an arithmetic sequence is 30 and the sum of the first eleven terms is 55. what is the common difference?
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the eleventh term of an arithmetic sequence is 30 and the sum of the first eleven terms is 55. what is the common difference?
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The value of Common Difference(d) = 5.
Step-by-step explanation:
★ Given that :
◼ Eleventh term of an AP series : 30
◼ Sum of first eleven terms of an AP series ; 55
★ To find :
◼ The common difference (d).
★ Let :
[tex]\sf \rightarrow a_{11} : a + 10d = 30\:.....(1)[/tex]
[tex]\sf \rightarrow S_{n} = \cfrac{n}{2} \bigg [ 2a + (n-1) d\bigg ][/tex]
[tex]\sf \implies 55 = \cfrac{11}{2} \bigg [ 2a + (11-1) d\bigg ][/tex]
[tex]\sf \implies \cancel{55} \times \cfrac{2}{\cancel{11}} = 2a + 10d[/tex]
[tex]\sf \implies 2a + 10d = 10[/tex]
[tex]\sf \implies 2(a + 5d) = 10[/tex]
[tex]\sf \implies a + 5d= \cancel{\cfrac{10}{2}}[/tex]
[tex]\sf \implies a + 5d=5 \:.....(2)[/tex]
[tex]\sf \implies 5d = 25[/tex]
[tex]\sf \implies d = \cancel{\cfrac{25}{5}}[/tex]
[tex]\sf \implies d = 5[/tex]
[tex]\sf \implies a + 10(5)=30[/tex]
[tex]\sf \implies a + 50=30[/tex]
[tex]\sf \implies a=30-50[/tex]
[tex]\sf \implies a=-20[/tex]
★ Verification :
Substitute the values of a & d in (1) to get LHS = RHS.
[tex]\sf \implies -20+10(5)[/tex]
[tex]\sf \implies -20+50[/tex]
[tex]\sf \implies -30[/tex]
☯ Since, LHS = RHS.
☯ Hence, it was verified.
[tex]\underline{\boxed{\rm{\purple{\therefore Value\:of\:Common\:difference(d)\:=5.}}}}\:\orange{\bigstar}[/tex]
★ More info :
Formulae related to Arithmetic Progression :-
[tex]\tt\red{\underline{\underline{\blacksquare \: a_{n} = a + (n -1)d}}}[/tex]
[tex]\tt\red{\underline{\underline{\blacksquare \: S_{n} = \cfrac{n}{2} \bigg[a + (n -1)d \bigg]}}}[/tex]
Where,
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