Question 59(a) Write the nth term (Tn) of an Arithmetic Progression (A.P.) consisting of all whole numbers which are divisible by 3 and 7.(b) How many of these are two-digit numbers? Write them.(c) Find the sum of first 10 terms of this A.P.

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(a) A.P. = 21, 42, 63, .........The above sequence is an A.P. with first term (a) = 21 and common difference (d) = 21.By formula,Tn = a + (n - 1)d= 21 + (n - 1)21= 21 + 21n - 21= 21n.Hence, the nth term = 21n.(b) A.P. = 21, 42, 63, 84, 105, ........Hence, there are four two digit numbers i.e. 21, 42, 63, 84 in the A.P.(c) By formula,Sum of A.P. = n2(a+an)\dfrac{n}{2}(a + a_n)2n​(a+an​)Sum of first 10 terms of A.P.=102(a+a10)=5[a+a+(n−1)d]=5[2a+(n−1)d]=5[2×21+(10−1)×21]=5[42+9×21]=5[42+189]=5×231=1155.\text{Sum of first 10 terms of A.P.} = \dfrac{10}{2}(a + a_{10}) \[1em] = 5[a + a + (n - 1)d] \[1em] = 5[2a + (n - 1)d] \[1em] = 5[2 \times 21 + (10 - 1) \times 21] \[1em] = 5[42 + 9 \times 21] \[1em] = 5[42 + 189] \[1em] = 5 \times 231 \[1em] = 1155.Sum of first 10 terms of A.P.=210​(a+a10​)=5[a+a+(n−1)d]=5[2a+(n−1)d]=5[2×21+(10−1)×21]=5[42+9×21]=5[42+189]=5×231=1155.Hence, sum of first 10 terms of the A.P. = 1155.

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Question 59