Answer :

Let Monday be the first term i.e. a = t1

Let Tuesday be the second term i.e t2


Let Wednesday be the third term i.e t3


Let Thursday be the fourth term i.e t4


Let Friday be the fifth term i.e t5


Let Saturday be the sixth term i.e t6


Given: t1 + t6 = 5 + (t2 + t6)


a = 5 + (t2 + t6) – t6


a = 5 + t2 …..(1)


We know that,


Now, By using nth term of an A.P. formula


tn = a + (n – 1)d


where n = no. of terms


a = first term


d = common difference


tn = nth terms


Thus, t2 = a + (2 – 1) d


t2 = a + d


Now substitute value of t2 in (1) we get,


a = 5 + (a + d)


d = a – 5 – a = – 5


Given: t3 = – 30°


Thus, t3 = a + (3 – 1) × ( – 5)


– 30 = a + 2 × ( – 5)


– 30 = a – 10


a = – 30 + 10 = – 20°


Thus, Monday, a = t1 = – 20°


Using formula tn + 1 = tn + d


We can find the value of the other terms


Tuesday, t2 = t1 + d = – 20 – 5 = – 25°


Wednesday, t3 = t2 + d = – 25 – 5 = – 30°


Thursday, t4 = t3 + d = – 30 – 5 = – 35°


Friday, t5 = t4 + d = – 35 – 5 = 40°


Saturday, t6 = t5 + d = – 40 – 5 = – 45°


Thus, we obtain an A.P.


– 20°, – 25°, – 30°, – 35°, – 40°, – 45°


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